86. 分隔链表https://leetcode-cn.com/problems/partition-list/
难度中等476
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。
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
通过次数123,981提交次数196,004
?
/**
* 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 first = new ListNode();
first.next = head;
ListNode temp = first;
ListNode node = first;
ListNode z;
while(temp.next!=null && temp.next.val<x)
{
temp = temp.next;
}
System.out.println(temp.val);
node = temp;
while(node.next!=null)
{
if(node.next.val<x)
{
z = node.next;
node.next = node.next.next;
z.next = temp.next;
temp.next = z;
temp = temp.next;
}
else node = node.next;
}
return first.next;
}
}
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