【算法一则】分隔链表

题目

给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有小于 x 的节点都出现在大于或等于 x 的节点之前。
在这里插入图片描述

你应当保留两个分区中每个节点的初始相对位置。

示例 1：

输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]
示例 2：输入：head = [2,1], x = 2
输出：[1,2]
提示：链表中节点的数目在范围 [0, 200] 内
-100 <= Node.val <= 100
-200 <= x <= 200

ListNode

public class ListNode {int val;ListNode next;public ListNode() {}ListNode(int x) {val = x;next = null;}ListNode(int x, ListNode next) {val = x;this.next = next;}
}

题解

package algorithm.link;import org.junit.Test;/*** Partition** @author allens* @date 2024/4/15*/
public class Partition {/*** partition** @param head* @param x* @return*/public ListNode partition(ListNode head, int x) {ListNode left = new ListNode();ListNode right = new ListNode();ListNode leftHead = left;ListNode rightHead = right;while (head != null) {if (head.val < x) {left.next = head;left = left.next;} else {right.next = head;right = right.next;}head = head.next;}right.next = null;left.next = rightHead.next;return leftHead.next;}@Testpublic void testMain () {ListNode head = new ListNode(1, new ListNode(4, new ListNode(3, new ListNode(2, new ListNode(5, new ListNode(2))))));ListNode partition = partition(head, 3);print(partition);System.out.println(partition);}private void print(ListNode partition) {while (partition != null) {System.out.print(partition.val + " ");partition = partition.next;}}}