19.2.23 [LeetCode 86] Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

题意

把链表按前面的全是小于x的，后面的全是大于等于x的顺序重组

题解

 1 class Solution {
 2 public:
 3     ListNode* partition(ListNode* head, int x) {
 4         ListNode*less = new ListNode(0),*lessh=less, *greater = new ListNode(0),*greaterh=greater;
 5         while (head) {
 6             ListNode*after = head->next;
 7             if (head->val < x) {
 8                 less->next = head;
 9                 less = less->next;
10                 less->next = NULL;
11             }
12             else {
13                 greater->next = head;
14                 greater = greater->next;
15                 greater->next = NULL;
16             }
17             head = after;
18         }
19         less->next = greaterh->next;
20         return lessh->next;
21     }
22 };
View Code

我就总爱各种地方有漏洞……面的时候也差不多……惨

posted @ 2019-02-23 09:37  TobicYAL  阅读(157)  评论(0)  编辑  收藏  举报