LeetCode-Partition List

最新推荐文章于 2021-02-23 21:38:10 发布

所难

最新推荐文章于 2021-02-23 21:38:10 发布

阅读量242

点赞数

CC 4.0 BY-SA版权

分类专栏： Cracking LeetCode List 文章标签： leetcode Partition List

本文链接：https://blog.youkuaiyun.com/HolyDominat/article/details/38438539

Cracking LeetCode 同时被 2 个专栏收录

152 篇文章

订阅专栏

List

17 篇文章

订阅专栏

本文介绍了一种链表分区算法，该算法将链表中小于给定值x的所有节点置于大于等于x的节点之前，同时保持各分区内部节点的原始相对顺序不变。例如，对于链表1->4->3->2->5->2 和 x = 3，算法返回1->2->2->4->3->5。文章提供了详细的解决方案和实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

Solution:

Code:

<span style="font-size:14px;">/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode *lessBegin = NULL, *lessEnd = NULL, *greaterBegin = NULL, *greaterEnd = NULL;
        while (head != NULL) {
            if (head->val < x) {
                if (lessBegin == NULL) {
                    lessBegin = head;
                    lessEnd = head;
                } else {
                    lessEnd->next = head;
                    lessEnd = head;
                }
                head = head->next;
                lessEnd->next = NULL;
            } else {
                if (greaterBegin == NULL) {
                    greaterBegin = head;
                    greaterEnd = head;
                } else {
                    greaterEnd->next = head;
                    greaterEnd = head;
                }
                head = head->next;
                greaterEnd->next = NULL;
            }
        }
        if (lessBegin != NULL) {
            lessEnd->next = greaterBegin;
            return lessBegin;
        } else 
            return greaterBegin;
    }
};</span>