合并两个排序的链表

最新推荐文章于 2025-07-20 12:45:35 发布

samshoot

最新推荐文章于 2025-07-20 12:45:35 发布

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CC 4.0 BY-SA版权

文章标签： c++ 算法 leetcode

本文链接：https://blog.youkuaiyun.com/c0c0nut/article/details/121577909

/*
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 || pHead2 == NULL)
           return pHead1 != NULL ? pHead1 : pHead2;
        ListNode* tmp = new ListNode(0);
        ListNode* newp =tmp;
        while(pHead1 && pHead2){
            if(pHead1->val < pHead2->val){
                tmp->next = pHead1;
                pHead1 = pHead1->next;
            }
            else{
                tmp->next = pHead2;
                pHead2 = pHead2->next;
            }
            tmp = tmp->next;
            if(pHead1)
                tmp->next = pHead1;
            if(pHead2)
                tmp->next = pHead2;
        }
        newp = newp->next;
        return newp;
    }
};

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

/*
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 || pHead2 == NULL)
           return pHead1 != NULL ? pHead1 : pHead2;
        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;
    }
};