《剑指offer》刷题笔记（代码的鲁棒性）：合并两个排序的链表

今天药忘吃喽~ 2022-06-07 08:18 107阅读 0赞

# 《剑指offer》刷题笔记（代码的鲁棒性）：合并两个排序的链表 #

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*  **转载请注明作者和出处：[http://blog.csdn.net/u011475210][http_blog.csdn.net_u011475210]**
 *  **代码地址：[https://github.com/WordZzzz/CodingInterviewChinese2][https_github.com_WordZzzz_CodingInterviewChinese2]**
 *  **文章地址：[https://github.com/WordZzzz/Note/tree/master/AtOffer][https_github.com_WordZzzz_Note_tree_master_AtOffer]**
 *  **刷题平台：[https://www.nowcoder.com/][https_www.nowcoder.com]**
 *  **题  库：剑指offer**
 *  **编  者：WordZzzz**

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*  剑指offer刷题笔记代码的鲁棒性合并两个排序的链表
    
     *  前言
     *  题目描述
     *  解题思路
     *  C版代码实现
        
         *  递归
         *  循环
     *  Python 代码实现
        
         *  递归
         *  循环

## 前言 ##

在面试过程中，最容易犯两种错误：一是在写代码之前没有对合并的过程想清楚，最终合并出来的链表要么中间断开了要么并没有做到递增排序；二是代码在鲁棒性方面存在问题，程序一旦有特殊的输入（如空链表）就会崩溃。

## 题目描述 ##

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

## 解题思路 ##

递归：

新创建一个指针就可以，比较两个链表的值，然后做相应递归更新。

循环：

需要创建两个指针，一个指向合并链表的表头，另一个用于更新，不断指向合并链表的表尾。最后返回指向表头的指针即可。

需要注意的是，我为了简化代码，新建的是指向带有头结点的链表（链表有带头结点和不带头结点点两种）。如果全部初始化为NULL（python是None），那么我在循环之前，就得加个判断来给合并链表的第一个结点赋值。

## C++版代码实现 ##

### 递归 ###

/*
    /*
    struct ListNode {
        int val;
        struct ListNode *next;
        ListNode(int x) :
                val(x), next(NULL) {
        }
    };*/
    class Solution {
    public:
        ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
        {
            //边界判断
            if(pHead1 == NULL)
                return pHead2;
            else if(pHead2 == NULL)
                return pHead1;
            //创建头指针
            ListNode* pMergeHead = NULL;
            if(pHead1->val < pHead2->val){
                pMergeHead = pHead1;
                pMergeHead->next = Merge(pHead1->next, pHead2);
            }
            else{
                pMergeHead = pHead2;
                pMergeHead->next = Merge(pHead1, pHead2->next);
            }
            return pMergeHead;
        }
    };

### 循环 ###

/*
    struct ListNode {
        int val;
        struct ListNode *next;
        ListNode(int x) :
                val(x), next(NULL) {
        }
    };*/
    class Solution {
    public:
        ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
        {
            //边界判断
            if(pHead1 == NULL)
                return pHead2;
            else if(pHead2 == NULL)
                return pHead1;
            //创建头尾指针
            ListNode* pMergeTail = new ListNode(0);
            ListNode* pMergeHead = new ListNode(0);
            //尾指针赋值
            pMergeTail = pMergeHead;
            //循环开始
            while(pHead1 && pHead2){
                if(pHead1->val < pHead2->val){
                    pMergeTail->next = pHead1;
                    pHead1 = pHead1->next;
                }
                else{
                    pMergeTail->next = pHead2;
                    pHead2 = pHead2->next;
                    }
                pMergeTail = pMergeTail->next;
            }
            //剩下的链表部分直接添加
            pMergeTail->next = pHead1 ? pHead1 : pHead2;
            return pMergeHead->next;
        }
    };

## Python 代码实现 ##

### 递归 ###

# -*- coding:utf-8 -*-
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    class Solution:
        # 返回合并后列表
        def Merge(self, pHead1, pHead2):
            # write code here
            if pHead1 is None:
                return pHead2
            elif pHead2 is None:
                return pHead1
    
            pMergeHead = ListNode(0)
    
            if pHead1.val < pHead2.val:
                pMergeHead = pHead1
                pMergeHead.next = self.Merge(pHead1.next, pHead2)
            else:
                pMergeHead = pHead2
                pMergeHead.next = self.Merge(pHead1, pHead2.next)
    
            return pMergeHead

### 循环 ###

# -*- coding:utf-8 -*-
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    class Solution:
        # 返回合并后列表
        def Merge(self, pHead1, pHead2):
            # write code here
            if pHead1 is None:
                return pHead2
            elif pHead2 is None:
                return pHead1
    
            pMergeTail = pMergeHead = ListNode(0)
    
            while pHead1 and pHead2:
                if pHead1.val < pHead2.val:
                    pMergeTail.next = pHead1
                    pHead1 = pHead1.next
                else:
                    pMergeTail.next = pHead2
                    pHead2 = pHead2.next
                pMergeTail = pMergeTail.next
    
            pMergeTail.next = pHead1 or pHead2
            return pMergeHead.next

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[http_blog.csdn.net_u011475210]: http://blog.csdn.net/u011475210
[https_github.com_WordZzzz_CodingInterviewChinese2]: https://github.com/WordZzzz/CodingInterviewChinese2
[https_github.com_WordZzzz_Note_tree_master_AtOffer]: https://github.com/WordZzzz/Note/tree/master/AtOffer
[https_www.nowcoder.com]: https://www.nowcoder.com/