圖解算法——合並兩個有序的列表（Merge Two Sorted Lists）

1. 題目描述

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

 

2.示例

示例1：

 

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]

示例2：

Input: l1 = [], l2 = []
Output: []

示例3：

Input: l1 = [], l2 = [0]
Output: [0]

3. 要求

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both l1 and l2 are sorted in non-decreasing order.

4.解題思想

和上篇博客 圖解算法——合並k個排序列表（Merge k Sorted Lists） 類似，

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    //非遞歸
    public ListNode mergeTwoLists(ListNode l1, ListNode l2){
        ListNode head = new ListNode(0);
        ListNode res = head;
        while(l1 != null && l2!=null){
            //res.val = l1.val>l2.val?l2.val:l1.val;
            if(l1.val>=l2.val){
                //res.val = l2.val;
                res.next = l1;//是res.next而不是賦值操作.val
　　　　　　　　　　l2 = l2.next;                
            }else if(l1.val<l2.val){
                //res.val = l1.val;
                res.next = l2;
　　　　　　　　　　l1 = l1.next;
            }
            res = res.next;
        }
        
        if(l1 != null){
            res.next = l1;//這里是res.next，不是res
        }
        if(l2 != null){
            res.next = l2;//這里是res.next，不是res
        }
        return head.next;//因為初始化時候設置初始節點為0，l1和l2的節點從第二個開始
    }
    
    //遞歸
    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;
        }
        
    }
}

 

 

 

Over......