【LeetCode】Partition List

最新推荐文章于 2024-01-17 14:19:03 发布

原创最新推荐文章于 2024-01-17 14:19:03 发布 · 539 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 2 个专栏收录

86 篇文章

订阅专栏

链表

12 篇文章

订阅专栏

本文详细阐述了如何通过链表实现元素分区，确保所有小于指定值的节点位于前面，大于等于该值的节点位于后面，同时保持原始相对顺序不变。

Partition List

Total Accepted: 14597 Total Submissions: 54894 My Submissions

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.

这类问题，使用两个链表思路会清晰

/**
 * 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) {
        if (!head || !(head->next)) return head;
        ListNode *head_left = NULL, *tail_left = NULL, *head_right = NULL, *tail_right = NULL;
        ListNode *p = head, *q = NULL;
        while (p) {
            q = p->next;
            if (p->val < x) {
                if(!head_left) {
                    head_left = p;
                } else {
                    tail_left->next = p;
                }
                tail_left = p;
                tail_left->next = NULL;
            } else {
                if(!head_right) {
                    head_right = p;
                } else {
                    tail_right->next = p;
                }
                tail_right = p;
                tail_right->next = NULL;
            }
            p = q;
        }
        if (tail_left) {/////////////////////////////////
            tail_left->next = head_right;
            return head_left;
        } else {
            return head_right;
        }
    }
};