leetcode 81: Partition List

最新推荐文章于 2024-11-19 23:58:32 发布

xudli

最新推荐文章于 2024-11-19 23:58:32 发布

阅读量1.8k

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/xudli/article/details/8587457

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，同时保持各分区内的相对顺序不变。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Partition ListApr 30 '12

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if( head==null) return head;
        
        ListNode less = new ListNode(-1);
        ListNode greater = new ListNode(-1);
        
        ListNode p=less, q=greater;
        
        while(head!=null) {
            if(head.val<x) {
                p.next = head;
                head = head.next;
                p = p.next;
                p.next = null;
            } else {
                q.next = head;
                head = head.next;
                q = q.next;
                q.next = null;
            }
            
        }
        
        p.next = greater.next;
        return less.next;
    }
}