[剑指Offer] 16.合并两个排序链表

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。

【思路1】递归

/*
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 == NULL)
            return pHead2;
        else if(pHead2 == NULL)
            return pHead1;
        ListNode* res = NULL;
        if(pHead1->val <= pHead2->val){
            res = pHead1;
            res->next = Merge(pHead1->next, pHead2);
        }else{
            res = pHead2;
            res->next = Merge(pHead1,pHead2->next);
        }
        return res;
    }
};

 【思路2】非递归，新建一个链表并保存头结点，将原来两个链表进行比较按顺序插入到新链表中，最后将有剩余的链表直接接上。

/*
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 == NULL)
            return pHead2;
        else if(pHead2 == NULL)
            return pHead1;
        ListNode* res = NULL;
        ListNode* cur = NULL;
        while(pHead1 != NULL && pHead2 != NULL){
            if(pHead1->val <= pHead2->val){
                if(res == NULL)
                    res = cur = pHead1;
                else{
                    cur->next = pHead1;
                    cur = cur->next;
                }
                pHead1 = pHead1->next;
            }else{
                if(res == NULL)
                    res = cur = pHead2;
                else{
                    cur->next = pHead2;
                    cur = cur->next;
                }
                pHead2 = pHead2->next;
            }
        }
        if(pHead1 == NULL){
            cur->next = pHead2;
        }
        if(pHead2 == NULL){
            cur->next = pHead1;
        }
        return res;
    }
};