LeetCode-JAVA-Partition List

最新推荐文章于 2025-09-11 19:19:55 发布

clevery017

最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： JAVA 文章标签： leetcode

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本文介绍了一种链表分隔算法，该算法可以将链表中小于特定值x的节点放置在大于等于x的节点之前，并保持各部分内部节点原有的相对顺序。通过创建两个临时链表来分别收集小于x和大于等于x的节点，最后将这两个链表连接起来返回结果。

题目：

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.

思路：

设定两个list，当值比x小时，放到list1中；比x大时，放到list2中。再将两个list合并。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode h1 = null;
        ListNode h2 = null;
        ListNode tmph1 = null, tmph2 = null;
        while(head!=null){
            ListNode h = new ListNode(head.val);
            if(head.val<x){
                if(h1 == null){
                    h1 = h;
                    tmph1 = h1;
                }else{
                    h1.next = h;
                    h1 = h1.next;
                }
            }else{
                if(h2==null){
                    h2 = h;
                    tmph2 = h2;
                }else{
                    h2.next = h;
                    h2=h2.next;
                }
            }
            head = head.next;
        }
        if(tmph1!=null){
            h1.next = tmph2;
        }else if(tmph2!=null){
            return tmph2;
        }
        return tmph1;
    }
}