Integer to Roman


/*
my implementation is kind of too complicated, mabye, just simply define 9 strings for each digit, much easier....
*/

/*
Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.
*/
public class Solution {
    public String intToRoman(int num) {
        String s = new String();
        char[][] cs={{'M', 'M', 'M'},  {'C', 'D', 'M'}, {'X', 'L', 'C'}, {'I', 'V', 'X'}};
        int i=0;
        for(int k=1000; k>=1; k=k/10)
        {
            int m = num/k;
            num = num >= k ? num%k : num;
            s += intToRoman(m, cs[i]);
            i++;
        }
        return s;
    }
    private String intToRoman(int num, char[] c)
    {
        if (num==0)
            return new String();
            
        String s=new String();
        
        if(num==9)
        {
            s += c[0];
            s += c[2];
        }
        else if(num==5)
            s += c[1];
        else if(num==4)
        {
            s += c[0];
            s += c[1];
        }
        else
        {
            if(num>5)
            {
                num = num%5;
                s += c[1];
            }
            
            for(int i=0; i<num; i++)
                s+=c[0];
        }
        return s;
    }
}
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ZigZag Conversion


/*
O(N)
*/

/*
The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
*/

public class Solution {
    public String convert(String s, int nRows) 
    {
        if(nRows<=1)
            return s;
        String ret = new String();
        int stp=(2*nRows-2);
        int l=s.length();
        for(int i=0; i<nRows; i++)
        {
            int k=i;
            while(k<l)
            {
                ret += s.charAt(k);
                if(i==0)
                {
                    k+=stp;
                    continue;
                }
                k += stp-2*i;
                if(k>=l)
                    break;
                if(stp!=2*i)
                    ret += s.charAt(k);
                k += 2*i;
            }
        }
        return ret;
    }
}

dfs of graph with matrix representation

Tags

,

//see: http://www.cs.cornell.edu/~wdtseng/icpc/notes/graph_part1.pdf

//dfs of graph (not DCG)
//let's use matrix
public class Graph
{
	//starts from a, dfs
	public void dfs(int a, int[][] m)
	{
		System.out.printf("%d=>", a);
		//use first row to represent whether it's visited
		m[0][a] = 1;
		for(int i=1; i<m[a].length; i++)
			if(m[0][i]==0 && m[a][i]==1)
				dfs(i, m);
	}
}


//test code
public class Test
{
	public static void main(String[] args)
	{
		Graph g=new Graph();
		//http://en.wikipedia.org/wiki/Graph_(mathematics)	
		int[][] m = new int[7][7];
		m[1][2] = 1;
		m[1][5] = 1;
		m[2][1] = 1;
		m[2][3] = 1;
		m[2][5] = 1;
		m[3][2] = 1;
		m[3][4] = 1;
		m[4][3] = 1;
		m[4][5] = 1;
		m[4][6] = 1;
		m[5][1] = 1;
		m[5][2] = 1;
		m[5][4] = 1;
		m[6][4] = 1;
		
		g.dfs(2, m);
	}
}

Sort List

Tags

/*
Sort a linked list in O(n log n) time using constant space complexity.
*/
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null || head.next==null)
            return head;
            
        int size=0;
        ListNode lh=head;
        ListNode lt=head;
        ListNode rt=head;//tail
        while(rt.next!=null)
        {
            rt=rt.next;
            size++;
            
            if(rt.next==null)
                break;
            rt=rt.next;
            size++;
            lt=lt.next;
        }
        ListNode rh = lt.next;
        lt.next = null;
        return mergeTwoLists(sortList(lh), sortList(rh));
    }
    
//below code is from another problem: merge two sorted list

        private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1==null && l2==null)
            return null;
        else if (l1==null)
            return l2;
        else if (l2==null)
            return l1;
        ListNode h=null;
        if(l1.val<l2.val)
        {
            h=l1;
            l1=l1.next;
        }
        else
        {
            h=l2;
            l2=l2.next;
        }
        ListNode c=h;
        while(l1!=null &&l2!=null)
        {
            if(l1.val<l2.val)
            {
                c.next=l1;
                l1=l1.next;
            }
            else
            {
                c.next=l2;
                l2=l2.next;
            }
            c = c.next;
        }
        if(l1!=null)
            c.next = l1;
        else
            c.next = l2;
        return h;
    }
}

redblack tree

Tags

/**
not my implementation, from princeton. it looks like its implementation is not very efficient. if you compare it w/ java's own library, you will see the big performance difference
**/

/*************************************************************************
 *  Compilation:  javac RBTree.java
 *  Execution:    java RBTree < input.txt
 *  Dependencies: StdIn.java System.out.java  
 *  Data files:   http://algs4.cs.princeton.edu/33balanced/tinyST.txt  
 *    
 *  A symbol table implemented using a left-leaning red-black BST.
 *  This is the 2-3 version.
 *
 *  Note: commented out assertions because DrJava now enables assertions
 *        by default.
 *
 *  % more tinyST.txt
 *  S E A R C H E X A M P L E
 *  
 *  % java RBTree < tinyST.txt
 *  A 8
 *  C 4
 *  E 12
 *  H 5
 *  L 11
 *  M 9
 *  P 10
 *  R 3
 *  S 0
 *  X 7
 *
 *************************************************************************/

package RBTree;

import java.io.InputStream;
import java.util.Scanner;
import java.util.NoSuchElementException;
import java.util.Queue;
import java.util.LinkedList;

public class RBTree<Key extends Comparable<Key>, Value> {

    private static final boolean RED   = true;
    private static final boolean BLACK = false;

    private Node root;     // root of the BST

    // BST helper node data type
    private class Node {
        private Key key;           // key
        private Value val;         // associated data
        private Node left, right;  // links to left and right subtrees
        private boolean color;     // color of parent link
        private int N;             // subtree count

        public Node(Key key, Value val, boolean color, int N) {
            this.key = key;
            this.val = val;
            this.color = color;
            this.N = N;
        }
    }

   /*************************************************************************
    *  Node helper methods
    *************************************************************************/
    // is node x red; false if x is null ?
    private boolean isRed(Node x) {
        if (x == null) return false;
        return (x.color == RED);
    }

    // number of node in subtree rooted at x; 0 if x is null
    private int size(Node x) {
        if (x == null) return 0;
        return x.N;
    } 


   /*************************************************************************
    *  Size methods
    *************************************************************************/

    // return number of key-value pairs in this symbol table
    public int size() { return size(root); }

    // is this symbol table empty?
    public boolean isEmpty() {
        return root == null;
    }

   /*************************************************************************
    *  Standard BST search
    *************************************************************************/

    // value associated with the given key; null if no such key
    public Value get(Key key) { return get(root, key); }

    // value associated with the given key in subtree rooted at x; null if no such key
    private Value get(Node x, Key key) {
        while (x != null) {
            int cmp = key.compareTo(x.key);
            if      (cmp < 0) x = x.left;
            else if (cmp > 0) x = x.right;
            else              return x.val;
        }
        return null;
    }

    // is there a key-value pair with the given key?
    public boolean contains(Key key) {
        return (get(key) != null);
    }

    // is there a key-value pair with the given key in the subtree rooted at x?
    // private boolean contains(Node x, Key key) {
    //    return (get(x, key) != null);
    // }

   /*************************************************************************
    *  Red-black insertion
    *************************************************************************/

    // insert the key-value pair; overwrite the old value with the new value
    // if the key is already present
    public void put(Key key, Value val) {
        root = put(root, key, val);
        root.color = BLACK;
        // assert check();
    }

    // insert the key-value pair in the subtree rooted at h
    private Node put(Node h, Key key, Value val) { 
        if (h == null) return new Node(key, val, RED, 1);

        int cmp = key.compareTo(h.key);
        if      (cmp < 0) h.left  = put(h.left,  key, val); 
        else if (cmp > 0) h.right = put(h.right, key, val); 
        else              h.val   = val;

        // fix-up any right-leaning links
        if (isRed(h.right) && !isRed(h.left))      h = rotateLeft(h);
        if (isRed(h.left)  &&  isRed(h.left.left)) h = rotateRight(h);
        if (isRed(h.left)  &&  isRed(h.right))     flipColors(h);
        h.N = size(h.left) + size(h.right) + 1;

        return h;
    }

   /*************************************************************************
    *  Red-black deletion
    *************************************************************************/

    // delete the key-value pair with the minimum key
    public void deleteMin() {
        if (isEmpty()) throw new NoSuchElementException("BST underflow");

        // if both children of root are black, set root to red
        if (!isRed(root.left) && !isRed(root.right))
            root.color = RED;

        root = deleteMin(root);
        if (!isEmpty()) root.color = BLACK;
        // assert check();
    }

    // delete the key-value pair with the minimum key rooted at h
    private Node deleteMin(Node h) { 
        if (h.left == null)
            return null;

        if (!isRed(h.left) && !isRed(h.left.left))
            h = moveRedLeft(h);

        h.left = deleteMin(h.left);
        return balance(h);
    }


    // delete the key-value pair with the maximum key
    public void deleteMax() {
        if (isEmpty()) throw new NoSuchElementException("BST underflow");

        // if both children of root are black, set root to red
        if (!isRed(root.left) && !isRed(root.right))
            root.color = RED;

        root = deleteMax(root);
        if (!isEmpty()) root.color = BLACK;
        // assert check();
    }

    // delete the key-value pair with the maximum key rooted at h
    private Node deleteMax(Node h) { 
        if (isRed(h.left))
            h = rotateRight(h);

        if (h.right == null)
            return null;

        if (!isRed(h.right) && !isRed(h.right.left))
            h = moveRedRight(h);

        h.right = deleteMax(h.right);

        return balance(h);
    }

    // delete the key-value pair with the given key
    public void delete(Key key) { 
        if (!contains(key)) {
            System.err.println("symbol table does not contain " + key);
            return;
        }

        // if both children of root are black, set root to red
        if (!isRed(root.left) && !isRed(root.right))
            root.color = RED;

        root = delete(root, key);
        if (!isEmpty()) root.color = BLACK;
        // assert check();
    }

    // delete the key-value pair with the given key rooted at h
    private Node delete(Node h, Key key) { 
        // assert contains(h, key);

        if (key.compareTo(h.key) < 0)  {
            if (!isRed(h.left) && !isRed(h.left.left))
                h = moveRedLeft(h);
            h.left = delete(h.left, key);
        }
        else {
            if (isRed(h.left))
                h = rotateRight(h);
            if (key.compareTo(h.key) == 0 && (h.right == null))
                return null;
            if (!isRed(h.right) && !isRed(h.right.left))
                h = moveRedRight(h);
            if (key.compareTo(h.key) == 0) {
                Node x = min(h.right);
                h.key = x.key;
                h.val = x.val;
                // h.val = get(h.right, min(h.right).key);
                // h.key = min(h.right).key;
                h.right = deleteMin(h.right);
            }
            else h.right = delete(h.right, key);
        }
        return balance(h);
    }

   /*************************************************************************
    *  red-black tree helper functions
    *************************************************************************/

    // make a left-leaning link lean to the right
    private Node rotateRight(Node h) {
        // assert (h != null) && isRed(h.left);
        Node x = h.left;
        h.left = x.right;
        x.right = h;
        x.color = x.right.color;
        x.right.color = RED;
        x.N = h.N;
        h.N = size(h.left) + size(h.right) + 1;
        return x;
    }

    // make a right-leaning link lean to the left
    private Node rotateLeft(Node h) {
        // assert (h != null) && isRed(h.right);
        Node x = h.right;
        h.right = x.left;
        x.left = h;
        x.color = x.left.color;
        x.left.color = RED;
        x.N = h.N;
        h.N = size(h.left) + size(h.right) + 1;
        return x;
    }

    // flip the colors of a node and its two children
    private void flipColors(Node h) {
        // h must have opposite color of its two children
        // assert (h != null) && (h.left != null) && (h.right != null);
        // assert (!isRed(h) &&  isRed(h.left) &&  isRed(h.right))
        //     || (isRed(h)  && !isRed(h.left) && !isRed(h.right));
        h.color = !h.color;
        h.left.color = !h.left.color;
        h.right.color = !h.right.color;
    }

    // Assuming that h is red and both h.left and h.left.left
    // are black, make h.left or one of its children red.
    private Node moveRedLeft(Node h) {
        // assert (h != null);
        // assert isRed(h) && !isRed(h.left) && !isRed(h.left.left);

        flipColors(h);
        if (isRed(h.right.left)) { 
            h.right = rotateRight(h.right);
            h = rotateLeft(h);
        }
        return h;
    }

    // Assuming that h is red and both h.right and h.right.left
    // are black, make h.right or one of its children red.
    private Node moveRedRight(Node h) {
        // assert (h != null);
        // assert isRed(h) && !isRed(h.right) && !isRed(h.right.left);
        flipColors(h);
        if (isRed(h.left.left)) { 
            h = rotateRight(h);
        }
        return h;
    }

    // restore red-black tree invariant
    private Node balance(Node h) {
        // assert (h != null);

        if (isRed(h.right))                      h = rotateLeft(h);
        if (isRed(h.left) && isRed(h.left.left)) h = rotateRight(h);
        if (isRed(h.left) && isRed(h.right))     flipColors(h);

        h.N = size(h.left) + size(h.right) + 1;
        return h;
    }


   /*************************************************************************
    *  Utility functions
    *************************************************************************/

    // height of tree (1-node tree has height 0)
    public int height() { return height(root); }
    private int height(Node x) {
        if (x == null) return -1;
        return 1 + Math.max(height(x.left), height(x.right));
    }

   /*************************************************************************
    *  Ordered symbol table methods.
    *************************************************************************/

    // the smallest key; null if no such key
    public Key min() {
        if (isEmpty()) return null;
        return min(root).key;
    } 

    // the smallest key in subtree rooted at x; null if no such key
    private Node min(Node x) { 
        // assert x != null;
        if (x.left == null) return x; 
        else                return min(x.left); 
    } 

    // the largest key; null if no such key
    public Key max() {
        if (isEmpty()) return null;
        return max(root).key;
    } 

    // the largest key in the subtree rooted at x; null if no such key
    private Node max(Node x) { 
        // assert x != null;
        if (x.right == null) return x; 
        else                 return max(x.right); 
    } 

    // the largest key less than or equal to the given key
    public Key floor(Key key) {
        Node x = floor(root, key);
        if (x == null) return null;
        else           return x.key;
    }    

    // the largest key in the subtree rooted at x less than or equal to the given key
    private Node floor(Node x, Key key) {
        if (x == null) return null;
        int cmp = key.compareTo(x.key);
        if (cmp == 0) return x;
        if (cmp < 0)  return floor(x.left, key);
        Node t = floor(x.right, key);
        if (t != null) return t; 
        else           return x;
    }

    // the smallest key greater than or equal to the given key
    public Key ceiling(Key key) {  
        Node x = ceiling(root, key);
        if (x == null) return null;
        else           return x.key;  
    }

    // the smallest key in the subtree rooted at x greater than or equal to the given key
    private Node ceiling(Node x, Key key) {  
        if (x == null) return null;
        int cmp = key.compareTo(x.key);
        if (cmp == 0) return x;
        if (cmp > 0)  return ceiling(x.right, key);
        Node t = ceiling(x.left, key);
        if (t != null) return t; 
        else           return x;
    }


    // the key of rank k
    public Key select(int k) {
        if (k < 0 || k >= size())  return null;
        Node x = select(root, k);
        return x.key;
    }

    // the key of rank k in the subtree rooted at x
    private Node select(Node x, int k) {
        // assert x != null;
        // assert k >= 0 && k < size(x);
        int t = size(x.left); 
        if      (t > k) return select(x.left,  k); 
        else if (t < k) return select(x.right, k-t-1); 
        else            return x; 
    } 

    // number of keys less than key
    public int rank(Key key) {
        return rank(key, root);
    } 

    // number of keys less than key in the subtree rooted at x
    private int rank(Key key, Node x) {
        if (x == null) return 0; 
        int cmp = key.compareTo(x.key); 
        if      (cmp < 0) return rank(key, x.left); 
        else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right); 
        else              return size(x.left); 
    } 

   /***********************************************************************
    *  Range count and range search.
    ***********************************************************************/

    // all of the keys, as an Iterable
    public Iterable<Key> keys() {
        return keys(min(), max());
    }

    // the keys between lo and hi, as an Iterable
    public Iterable<Key> keys(Key lo, Key hi) {
        Queue<Key> queue = new LinkedList<>();
        // if (isEmpty() || lo.compareTo(hi) > 0) return queue;
        keys(root, queue, lo, hi);
        return queue;
    } 

    // add the keys between lo and hi in the subtree rooted at x
    // to the queue
    private void keys(Node x, Queue<Key> queue, Key lo, Key hi) { 
        if (x == null) return; 
        int cmplo = lo.compareTo(x.key); 
        int cmphi = hi.compareTo(x.key); 
        if (cmplo < 0) keys(x.left, queue, lo, hi); 
        if (cmplo <= 0 && cmphi >= 0) queue.add(x.key);
        if (cmphi > 0) keys(x.right, queue, lo, hi); 
    } 

    // number keys between lo and hi
    public int size(Key lo, Key hi) {
        if (lo.compareTo(hi) > 0) return 0;
        if (contains(hi)) return rank(hi) - rank(lo) + 1;
        else              return rank(hi) - rank(lo);
    }


   /*************************************************************************
    *  Check integrity of red-black BST data structure
    *************************************************************************/
    private boolean check() {
        if (!isBST())            System.out.println("Not in symmetric order");
        if (!isSizeConsistent()) System.out.println("Subtree counts not consistent");
        if (!isRankConsistent()) System.out.println("Ranks not consistent");
        if (!is23())             System.out.println("Not a 2-3 tree");
        if (!isBalanced())       System.out.println("Not balanced");
        return isBST() && isSizeConsistent() && isRankConsistent() && is23() && isBalanced();
    }

    // does this binary tree satisfy symmetric order?
    // Note: this test also ensures that data structure is a binary tree since order is strict
    private boolean isBST() {
        return isBST(root, null, null);
    }

    // is the tree rooted at x a BST with all keys strictly between min and max
    // (if min or max is null, treat as empty constraint)
    // Credit: Bob Dondero's elegant solution
    private boolean isBST(Node x, Key min, Key max) {
        if (x == null) return true;
        if (min != null && x.key.compareTo(min) <= 0) return false;
        if (max != null && x.key.compareTo(max) >= 0) return false;
        return isBST(x.left, min, x.key) && isBST(x.right, x.key, max);
    } 

    // are the size fields correct?
    private boolean isSizeConsistent() { return isSizeConsistent(root); }
    private boolean isSizeConsistent(Node x) {
        if (x == null) return true;
        if (x.N != size(x.left) + size(x.right) + 1) return false;
        return isSizeConsistent(x.left) && isSizeConsistent(x.right);
    } 

    // check that ranks are consistent
    private boolean isRankConsistent() {
        for (int i = 0; i < size(); i++)
            if (i != rank(select(i))) return false;
        for (Key key : keys())
            if (key.compareTo(select(rank(key))) != 0) return false;
        return true;
    }

    // Does the tree have no red right links, and at most one (left)
    // red links in a row on any path?
    private boolean is23() { return is23(root); }
    private boolean is23(Node x) {
        if (x == null) return true;
        if (isRed(x.right)) return false;
        if (x != root && isRed(x) && isRed(x.left))
            return false;
        return is23(x.left) && is23(x.right);
    } 

    // do all paths from root to leaf have same number of black edges?
    private boolean isBalanced() { 
        int black = 0;     // number of black links on path from root to min
        Node x = root;
        while (x != null) {
            if (!isRed(x)) black++;
            x = x.left;
        }
        return isBalanced(root, black);
    }

    // does every path from the root to a leaf have the given number of black links?
    private boolean isBalanced(Node x, int black) {
        if (x == null) return black == 0;
        if (!isRed(x)) black--;
        return isBalanced(x.left, black) && isBalanced(x.right, black);
    } 


   /*****************************************************************************
    *  Test client
     * @param args
    *****************************************************************************/
    public static void main(String[] args) { 
        RBTree<String, Integer> st = new RBTree<>();
//        Scanner scanner = new Scanner(System.in);
//        String input = scanner.nextLine();
//        for (int i = 0; !input.isEmpty(); i++)
//        {
//            String key = input;
//            st.put(key, i);
//            input = scanner.nextLine();
//        }
        
        String[] inputs = {"S", "E", "A", "R", "C", "H", "X", "M", "P", "L"};
        for (int i = 0; i< inputs.length; i++)
        {
            String key = inputs[i];
            st.put(key, i);
        }
        for (String s : st.keys())
            System.out.println(s + " " + st.get(s));
        System.out.println();
    }
}

Max Binary Heap

Tags


/*
see:
clrs: ch6: HeapSort
c++: priority_queue (default is max)
java: PriorityQueue (default is min)
*/
class MaxBinHeap
{
	//let's use simple array to avoid extra performance penalty
	private int[] hp;
	private int l;
	public void build(int[] a)
	{
		l = a.length;
		hp = new int[l];
		System.arraycopy(a, 0, hp, 0, l);
		for(int i=l/2-1; i>=0; i--)
			heapify(i);
	}
	
	public int findKthMax(int k)
	{
		int rt = 0;
		for(int i=0; i<k; i++)
			rt = extractmax();
		return rt;
	}
	
	//extract max item
	public int extractmax()
	{
		if(l<1)
			throw new IllegalArgumentException("bad input!");
		int r = hp[0];
		hp[0] = hp[l-1];
		l--;
		heapify(0);
		return r;
	}
	
	//heapify element i
	private void heapify(int i)
	{
		int il = 2*i+1;
		int ir = 2*(i+1);
		int ilargest = 0;
		if(il<l && hp[il]>hp[i])
			ilargest = il;
		else
			ilargest = i;
		if(ir<l && hp[ir]>hp[ilargest])
			ilargest = ir;
		if(ilargest != i)
		{
			swap(ilargest, i);
			heapify(ilargest);
		}
		return;
	}
	
	private void swap(int a, int b)
	{
		int t = hp[a];
		hp[a] = hp[b];
		hp[b] = t;
	}
}

Merge k Sorted Lists

Tags


/*
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity
T(N)=O(NlgN)
*/
========================c++============================
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    struct compare
    {
        bool operator()(ListNode* l, ListNode* r)
        {
            return l->val > r->val;
        }
    };
    
    ListNode *mergeKLists(vector<ListNode *> &lists) 
    {
        priority_queue<ListNode*, vector<ListNode*>, compare > listT;
        //init
        for(size_t i=0; i<lists.size(); i++)
        {
            if(lists[i])
                listT.push(lists[i]);
        }
        
        ListNode h(0);
        ListNode* c=&h;
        while(!listT.empty())
        {
            ListNode* n = listT.top();
            listT.pop();
            if(n->next)
                listT.push(n->next);
            c->next = n;
            c=c->next;
        }
        return h.next;
    }
};

/*
================================java====================================
*/
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        if(lists.size()==0)
            return null;
        PriorityQueue<ListNode> q=new PriorityQueue<ListNode>(lists.size(), 
            new Comparator<ListNode>()
            {
                public int compare(ListNode l, ListNode r)
                {
                    if(l.val>r.val)
                        return 1;
                    else if(l.val<r.val)
                        return -1;
                    else
                        return 0;
                }
            }
            );
        
        for(ListNode n:lists)    
            if(n!=null)
                q.add(n);
            
        ListNode h=new ListNode(0);
        ListNode c=h;
        while(q.size()>0)
        {
            ListNode n=q.poll();
            if(n.next != null)
                q.add(n.next);
            c.next=n;
            c=c.next;
        }
        return h.next;
    }
}

Merge Two Sorted Lists


/*
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists

T(N)=O(N)
*/

//==================c++====================================

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if(!l1 &amp;amp;&amp;amp; !l2)
            return NULL;
        else if(!l1)
            return l2;
        else if(!l2)
            return l1;
        
        ListNode* head = NULL;
        if(l1-&amp;gt;val &amp;lt;= l2-&amp;gt;val)
        {
            head = l1;
            l1=l1-&amp;gt;next;
        }
        else
        {
            head = l2;
            l2=l2-&amp;gt;next;
        }
        
        ListNode* c=head;
        while(l1&amp;amp;&amp;amp;l2)
        {
            if(l1-&amp;gt;val &amp;lt;= l2-&amp;gt;val)
            {
                insert(c, l1);
                l1=l1-&amp;gt;next;
            }
            else
            {
                insert(c, l2);
                l2=l2-&amp;gt;next;
            }
        }
        if(l1)
            insert(c, l1);
        else
            insert(c, l2);
        
        return head;
    }
    
    void insert(ListNode*&amp;amp; c, ListNode* n)
    {
        if(c-&amp;gt;next != n)
            c-&amp;gt;next = n;
        c = n;
    }
};


//=================================java==========================
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1==null && l2==null)
            return null;
        else if (l1==null)
            return l2;
        else if (l2==null)
            return l1;
        ListNode h=null;
        if(l1.val<l2.val)
        {
            h=l1;
            l1=l1.next;
        }
        else
        {
            h=l2;
            l2=l2.next;
        }
        ListNode c=h;
        while(l1!=null &&l2!=null)
        {
            if(l1.val<l2.val)
            {
                c.next=l1;
                l1=l1.next;
            }
            else
            {
                c.next=l2;
                l2=l2.next;
            }
            c = c.next;
        }
        if(l1!=null)
            c.next = l1;
        else
            c.next = l2;
        return h;
    }
}