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leetcode 21 -- Merge Two Sorted Lists

时间:2017-07-24 13:23:09      阅读:137      评论:0      收藏:0      [点我收藏+]

标签:dcl   and   ring   nod   block   tno   ted   view   c++   

Merge Two Sorted Lists

题目:
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.


题意:
合并两个排序过的链表


思路:
这是个非经常见的问题,须要注意的就是给定两个链表是否为空。


代码:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1 == NULL && l2 == NULL){
            return NULL;
        }else if(l1 == NULL && l2 != NULL){
            return l2;
        }else if(l1 != NULL && l2 == NULL){
            return l1;
        }
        //指定头节点
        ListNode *new_head = NULL;
        if(l1->val < l2->val){
            new_head = l1;
            l1 = l1->next;
        }else{
            new_head = l2;
            l2 = l2->next;
        }
        ListNode *p = new_head;
        while(l1 != NULL && l2 != NULL){
            if(l1->val < l2->val){
                p->next = l1;
                p = p->next;
                l1 = l1->next;
            }else{
                p->next = l2;
                p = p->next;
                l2 = l2->next;
            }
        }
        //接上余下的链表段
        if(l1 != NULL){
            p->next = l1;
        }else{
            p->next = l2;
        }
        return new_head;
    }
};

leetcode 21 -- Merge Two Sorted Lists

标签:dcl   and   ring   nod   block   tno   ted   view   c++   

原文地址:http://www.cnblogs.com/jzdwajue/p/7227903.html

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