C++实现合并两个排序的链表

时间:2021-10-24 05:46:44

本文实例为大家分享了C++合并两个排序的链表,供大家参考,具体内容如下

问题描述

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。

?
1
2
3
4
5
6
7
struct ListNode {
 int val;
 struct ListNode *next;
 ListNode(int x) :
 val(x), next(NULL) {
 }
};

方法一

?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
class Solution {
public:
 ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
 {
  ListNode* newList = NULL; //新链表头
  ListNode* newListRear = NULL; //新链表尾
  // 先处理某个链表为空的情形
  if (pHead1 == NULL){
   return pHead2;
  }
  if (pHead2 == NULL){
   return pHead1;
  }
  // 把数值小的结点放入新链表,生成头节点
  if (pHead1->val <= pHead2->val){
   newList = pHead1;
   newListRear = pHead1;
   pHead1 = pHead1->next;
  }else{
   newList = pHead2 ;
   newListRear = pHead2;
   pHead2 = pHead2->next;
  }
  // 两表均不空的情形下,遍历
  while (pHead1 != NULL && pHead2 != NULL) {
   if (pHead1->val <= pHead2->val) {
    newListRear->next =pHead1;
    newListRear = pHead1;
    pHead1 = pHead1->next;
   }else{
    newListRear->next =pHead2;
    newListRear = pHead2;
    pHead2 = pHead2->next;
   }
  }
  //某一表为空时,把另一表接入新表表尾
  if (pHead1 == NULL) {
   newListRear->next = pHead2;
  }
  if (pHead2 == NULL) {
   newListRear->next = pHead1;
  }
  
  return newList;
 }
};

方法二(递归思想)

?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
 ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
 {
  if (pHead1 == NULL){
   return pHead2;
  }
  if (pHead2 == NULL){
   return pHead1;
  }
  
  if (pHead1->val <= pHead2->val){ // pHead1为合并后的头节点
   pHead1->next = Merge(pHead1->next, pHead2);
   return pHead1;
  }else{ // pHead2 为合并后的头节点
   pHead2->next = Merge(pHead1, pHead2->next);
   return pHead2;
  }
 }
};

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持服务器之家。

原文链接:https://blog.csdn.net/francis_xd/article/details/82803905