题目地址

https://leetcode-cn.com/problems/partition-list/

题目描述

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。



示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

题解

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode smallHead = new ListNode();
        ListNode largeHead = new ListNode();

        ListNode smallTail = smallHead;
        ListNode largeTail = largeHead;
        while (head != null) {
            if (head.val >= x) {
                // head 的值大于等于 x，追加到 large 链表
                largeTail.next = head;
                largeTail = largeTail.next;
            } else {
                // head 的值小于 x，追加到 small 链表
                smallTail.next = head;
                smallTail = smallTail.next;
            }
            head = head.next;
        }
        
        // 合并两个链表
        smallTail.next = largeHead.next;
        largeTail.next = null;

        return smallHead.next;
    }

}