合并两个排序的链表（C++实现）

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;
        }
    }
};