【剑指offer刷题】-25-合并两个排序的链表

难度:简单

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。

注:考察链表,是面试常见题

思路:递归思想(源于剑指offer)


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

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转载自blog.csdn.net/xmc_muncy/article/details/80736639