LeetCode 21. Merge Two Sorted Lists

原题链接在这里：https://leetcode.com/problems/merge-two-sorted-lists/

题目：

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.

题解：

Method 1:

recursion, stop condition: 一条list 空了，返回另外一条。否则比较两个node value, 取小的，对应的点后移，next 指向用作下次比较返回的结果。

Time Complexity: O(m+n), m 是list1的长度，n 是list2 的长度.

Space: recursion 用的stack的大小, O(m+n).

Method 2:

Iteration, 生成一个dummy, 对应生成currentNode 用来iterate. 每次比较两个点，小的连在current 的后面直到 一个为 null, 再把没iterate完的那个连在current后面即可.

Time Complexity: O(m+n).

Space O(1). 

AC 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) {
        /*
        //Method 1, Recursion
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        
        ListNode mergeHead;
        if(l1.val <= l2.val){
            mergeHead = l1;
            mergeHead.next = mergeTwoLists(l1.next,l2);
        }else{
            mergeHead = l2;
            mergeHead.next = mergeTwoLists(l1,l2.next);
        }
        
        return mergeHead;
        */
        
        //Method 2, Iteration
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        while(l1!=null && l2!= null){
            if(l1.val<=l2.val){
                current.next = l1;
                l1 = l1.next;
                current = current.next;
            }else{
                current.next = l2;
                l2 = l2.next;
                current = current.next;
            }
        }
        if(l1 != null){
            current.next = l1;
        }
        if(l2 != null){
            current.next = l2;
        }
        return dummy.next;
    }
}

AC JavaScript:

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function(l1, l2) {
    var dummy = {
        val : -1,
        next : null
    };
    
    var cur = dummy;
    while(l1 && l2){
        if(l1.val < l2.val){
            cur.next = l1;
            l1 = l1.next;
        }else{
            cur.next = l2;
            l2 = l2.next;
        }
        
        cur = cur.next;
    }
    
    cur.next = l1 || l2;
    return dummy.next;
};

跟上 Merge k Sorted Lists, Merge Sorted Array.