LeetCode Merge Two Sorted Lists（021）解法总结

描述

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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

思路

使用循环遍历的方式，将两个链表合并成为一个链表。每次合并之前都进行一次比较操作。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        //使用-1作为头部节点，返回链表为头部节点的下一个节点
        ListNode out = new ListNode(-1);
        ListNode cur = out;
        while(l1 != null && l2 != null){
            //比较并创建节点，加入链表
            ListNode new_node;
            if(l1.val < l2.val){
                new_node = new ListNode(l1.val);
                l1 = l1.next;
            }else{
                new_node = new ListNode(l2.val);
                l2 = l2.next;
            }
            cur.next = new_node;
            cur = cur.next;
        }
        //将剩下的节点拼接在链表后面
        if(l1 != null){
            cur.next = l1;
        }else{
            cur.next = l2;
        }
        return out.next;
    }
}

另一种思路

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2){
        //如果有个链表为空就拼接另一个链表
        if(l1 == null) return l2;
	if(l2 == null) return l1;
        //否则就递归拼接较小的那个链表节点
	if(l1.val < l2.val){
	    l1.next = mergeTwoLists(l1.next, l2);
	    return l1;
	} else{
	    l2.next = mergeTwoLists(l1, l2.next);
	    return l2;
	}
    }
}