96题 Partition List

最新推荐文章于 2024-02-03 21:19:22 发布

原创最新推荐文章于 2024-02-03 21:19:22 发布 · 81 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#链表 #算法 #面试

算法专栏收录该内容

232 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，同时保持各分区内的相对顺序不变。

Partition List

Description
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.

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: The first node of linked list
     * @param x: An integer
     * @return: A ListNode
     */
    public ListNode partition(ListNode head, int x) {
        // write your code here
        if(head == null ){
            return null ;
        }
        ListNode dummy1 = new ListNode(-1) ;
        ListNode dummy2 = new ListNode(-1) ;
        ListNode left = dummy1 , temp = dummy2 ;
         
       
        while(head != null){
            if(head.val < x){
                left.next = head ;
                left = left.next ;
            }else{
                temp.next = head ;
                temp = temp.next ;
            }
            head = head.next ;
        }
        temp.next = null ;
        left.next = dummy2.next ;
        return dummy1.next ;
        

    }
}