题目描述

解题思路

• 使用两个链表（small，large）分别存储小于x和大于x的节点
• 遍历链表，将小于x的节点插入small中，其他的插入large中
• 连接small和large，并返回。

代码实现

class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode small = new ListNode(0);
        ListNode smallHead = small;    //记录small头
        ListNode large = new ListNode(0);
        ListNode largeHead = large;    //记录large头

        while(head != null){
            if(head.val < x){    //插入small尾
                small.next = head;
                small = small.next;
            }else{                //插入large尾
                large.next = head;
                large = large.next;
            }
            head = head.next;
        }
        large.next = null;
        small.next = largeHead.next;    //拼接small和large
        return smallHead.next;
    }
}