练习：合并两个排序链表

练习：合并两个排序链表

一、题目要求：

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。

二、代码（2018–8-10）

/*
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* p1 = pHead1, *p2 = pHead2;
        ListNode* h = new ListNode(-1);
        ListNode* p = h;
        while(p1 && p2) {
            if(p1->val < p2->val) {
                p->next = p1;
                p = p->next;
                p1 = p1->next;
            }
            else {
                p->next = p2;
                p = p->next;
                p2 = p2->next;
            }
        }
        p->next = p1 ? p1 : p2;
        ListNode* head = h->next;
        delete h;
        return head;
    }
};

代码（2018-01-16）

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

说明

在创建链表、删除链表、增加节点、删除节点时候，创建一个头节点可以方便操作。