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。

struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { }};

方法一

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）

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; } }};

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