LeetCode 86. 分隔链表(Partition List)

Posted 2021-12-15 hglibin

86. 分隔链表
86. 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。

LeetCode86. Partition List中等

示例:

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

Java 实现
ListNode Class

public class ListNode 
    int val;
    ListNode next;

    ListNode(int x) 
        val = x;
    

    @Override
    public String toString() 
        return val + "->" + next;
    

方法一

import java.util.LinkedList;
import java.util.Queue;

class Solution 
    public ListNode partition(ListNode head, int x) 
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        Queue<Integer> queue = new LinkedList<>();
        while (head != null) 
            if (head.val < x) 
                tail.next = head;
                tail = tail.next;
             else 
                queue.add(head.val);
            
            head = head.next;
        
        while (!queue.isEmpty()) 
            tail.next = new ListNode(queue.poll());
            tail = tail.next;
        
        return dummy.next;
    

方法二

class Solution 
    public ListNode partition(ListNode head, int x) 
        ListNode dummy1 = new ListNode(0);
        ListNode dummy2 = new ListNode(0);
        ListNode curr1 = dummy1;
        ListNode curr2 = dummy2;
        while (head != null) 
            if (head.val < x) 
                curr1.next = head;
                curr1 = curr1.next;
             else 
                curr2.next = head;
                curr2 = curr2.next;
            
            head = head.next;
        
        curr2.next = null;  // very important!
        curr1.next = dummy2.next;
        return dummy1.next;
    

参考资料

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