- 转载请注明作者和出处:http://blog.youkuaiyun.com/u011475210
- 代码地址:https://github.com/WordZzzz/Note/tree/master/LeetCode
- 刷题平台:https://www.nowcoder.com/ta/leetcode
- 题 库:Leetcode经典编程题
- 编 者:WordZzzz
题目描述
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.
解题思路
这道题剑指offer上应该也是有的,我就不去翻了。既然已经排序了,事情就变得简单多了,自下而上的归并排序的最后一步嘛~我们还是申请个新的结点,然后直接遍历两个链表元素比大小就好。
需要注意的是循环的终止条件,循环结束后还要把剩下的链表追加到cur的屁股后面哦。
C++版代码实现
/**
* 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)
return l2;
if(l2 == NULL)
return l1;
ListNode *dummy = new ListNode(0);
ListNode *cur = dummy;
while(l1 != NULL && l2 != NULL){
if(l1->val < l2->val){
cur->next = l1;
l1 = l1->next;
}
else{
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
cur->next = l1 ? l1 : l2;
return dummy->next;
}
};
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完的汪(∪。∪)。。。zzz