合并两个排序的链表

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。

 

非递归版本：

1.新建一个头结点mergeHead,和尾节点 current.

2.比较两个链表的头节点，确定mergeHead, 将current指向mergeHead.

3.依次比较两个链表节点（在都不为空的情况下），current->next保存数值小的节点, current后移

4.若链表为空, current->next指向 非空链表

5.返回mergeHead.

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
    if(pHead1 == nullptr)
        return pHead2;
    if(pHead2 == nullptr)
        return pHead1;
    
    ListNode * mergeHead = nullptr; //新建一个头结点mergeHead,和尾节点 current.
    ListNode * current = nullptr;
    if(pHead1->val <= pHead2->val)  //比较两个链表的头节点，确定mergeHead
    {                                //将current指向mergeHead.
        mergeHead =  current =pHead1;
        pHead1 = pHead1->next;
    }
    else
    {
        mergeHead = current = pHead2;
        pHead2 = pHead2->next;
    }
    
    while(pHead1!=NULL && pHead2!=NULL) //遍历两个链表
    {
        if(pHead1->val <= pHead2->val)  
        {
            current->next = pHead1;  //current保存小的节点
            current = current->next; //current后移
            pHead1 = pHead1->next;   //该链表后移
        }
        else
        {
            current->next = pHead2;
            current = current->next;
            pHead2 = pHead2->next;
        }
    }
    if(pHead1 == NULL)
    {
        current->next = pHead2;
    }
    if(pHead2 == NULL)
    {
        current->next = pHead1;
    }
    return mergeHead;   
    }
};