【题目】
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2
and x = 3,
return 1->2->2->4->3->5
.
【解析】
题意:给定一个单链表和一个x,把链表中小于x的放到前面,大于等于x的放到后面,每部分元素的原始相对位置不变。
思路:其实很简单,遍历一遍链表,把小于x的都挂到head1后,把大于等于x的都放到head2后,最后再把大于等于的链表挂到小于链表的后面就可以了。
这道题不难,主要是给不熟悉指针的同学学习交流。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode partition(ListNode head, int x) { ListNode lessHead = new ListNode(0); ListNode greaterHead = new ListNode(0); ListNode node = head, less = lessHead, greater = greaterHead; while (node != null) { ListNode next = node.next; if (node.val < x) { less.next = node; less = less.next; less.next = null; } else { greater.next = node; greater = greater.next; greater.next = null; } node = next; } less.next = greaterHead.next; return lessHead.next; } }
时间: 2024-10-09 02:36:15