剑指Offer-16-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。

Code

//非递归版
/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    //归并排序：升序
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode* result= new ListNode(-1);
        ListNode* current = result;
        while(pHead1!=0&&pHead2!=0){
            if(pHead1->val<pHead2->val){
                current->next = pHead1;
                pHead1=pHead1->next;
            }else{
                current->next = pHead2;
                pHead2=pHead2->next;
            }
            current=current->next;
        }
        if(pHead1!=0)
            current->next = pHead1;
        
        if(pHead2!=0)
            current->next = pHead2;
            
        //返回的链表不包括头节点
        return result->next; 
    }
};

//递归版
/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    //归并排序：升序
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        
        if(pHead1==0)return pHead2;
        if(pHead2==0)return pHead1;
        
        if(pHead1->val<pHead2->val){
            pHead1->next = Merge(pHead1->next,pHead2);
            return pHead1;
        }else{
            pHead2->next = Merge(pHead1,pHead2->next);
            return pHead2;
        }
    }
};