LeetCode21.-Merge Two Sorted Lists-Easy

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

解法1: 递归

https://mp.weixin.qq.com/s/7Tqc84twkk3bRmIebxwWAg

自上而下思考，递归就是将问题向下分解成若干相同子问题，直到子问题规模最小，无法继续分解，即base case，求得最小问题的解后，再依次向上归。

递归两个两个条件：1）子问题调用同一个递归函数 2）递归终止条件（递到最小子问题，返回，向上归）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        
        if(l1 == null) {
            return l2;
        }
        if(l2 == null) {
            return l1;
        }
        
        if(l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }
}

解法2: 

新建一个链表，需要两个Node，一个始终指向头结点，另一随着链表的增长而移动，始终指向链表最后。

每次插入当前l1 l2中最小值。

当一个链表遍历完后，另一个未完成的链表直接链入新链表的末尾。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        
        ListNode result = new ListNode(0);
        ListNode cur = result;
        
        while(l1 != null && l2 != null) {
            if(l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        
        cur.next = (l1 == null) ? l2 : l1;
        
        return result.next;
    }
}