剑指offer16：输入两个单调递增的链表，合成后的链表满足单调不减规则。

1 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。

2 思路与方法

　　迭代法：两个链表中较小的头结点作为合并后头结点，之后依次合并两个链表中较小的结点，以此类推，最终合并剩余结点； ListNode* out_list =s->Merge(l1,l4)；

　　递归法：两个链表中较小的头结点作为合并后头结点，递归；（不推荐递归）

3 C++核心代码

3.1 迭代实现

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode* phead = new ListNode(0);
        ListNode* list_new = phead;
        while(pHead1 || pHead2){
            if(pHead1==NULL){
                list_new->next=pHead2;
                break;
            }
            else if(pHead2==NULL){
                list_new->next = pHead1;
                break;
            }
            if((pHead1->val)>(pHead2->val))
            {
                list_new->next = pHead2;
                list_new = list_new->next;
                pHead2=pHead2->next;
            }
            else{
                list_new->next = pHead1;
                list_new = list_new->next;
                pHead1=pHead1->next;
            }
        }
        return phead->next;
    }
};

3.2 递归实现

/*
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)
            return pHead2;
        if (!pHead2)
            return pHead1;
 
        if (pHead1->val < pHead2->val)
            pHead1->next = Merge(pHead1->next, pHead2);
        else
            pHead2->next = Merge(pHead1, pHead2->next);
 
        return pHead1->val < pHead2->val ? pHead1 : pHead2;
    }
};

4 完整代码

#include <iostream>

using namespace std;


struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode* phead = new ListNode(0);
        ListNode* list_new = phead;
        while (pHead1 || pHead2){
            if (pHead1 == NULL){
                list_new->next = pHead2;
                break;
            }
            else if (pHead2 == NULL){
                list_new->next = pHead1;
                break;
            }
            if ((pHead1->val)>(pHead2->val))
            {
                list_new->next = pHead2;
                list_new = list_new->next;
                pHead2 = pHead2->next;
            }
            else{
                list_new->next = pHead1;
                list_new = list_new->next;
                pHead1 = pHead1->next;
            }
        }
        return phead->next;
    }
};
int main()
{
    Solution *s = new Solution();
    //vector<int> v = { 2,4,6,1,3,5,7 };
    ListNode *l1 = new ListNode(1);
    ListNode *l2 = new ListNode(6);
    ListNode *l3 = new ListNode(8);
    ListNode *l4 = new ListNode(4);
    ListNode *l5 = new ListNode(5);
    ListNode *l6 = new ListNode(6);
    ListNode *l7 = new ListNode(7);
    l1->next = l2;
    l2->next = l3;
    //l3->next = l4; 
    l4->next = l5;
    l5->next = l6;
    l6->next = l7;

    ListNode* out_list = s->Merge(l1, l4);
    while (out_list){
        cout << out_list->val << " ";
        out_list = out_list->next;
    }
    cout << endl;

    system("pause");
    return 0;
}

参考资料

https://blog.csdn.net/ansizhong9191/article/details/80697615