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.
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/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */public
class Solution { /**This is a fundamental problem. But we need to pay attention to some subtle problems. * such as the null pointers. * @author Averill Zheng * @version 2016-06-07 * @since JDK 1.7 */ public
ListNode partition(ListNode head, int
x) { ListNode newHead = new
ListNode(0); if(head != null){ newHead.next = head; ListNode firstLarger = head; ListNode insertPosition = newHead; while(firstLarger != null
&& firstLarger.val < x){ insertPosition = insertPosition.next; firstLarger = firstLarger.next; } ListNode current = firstLarger; while(firstLarger != null){ if(firstLarger.val >= x){ current = firstLarger; firstLarger = firstLarger.next; } else{ current.next = firstLarger.next; firstLarger.next = insertPosition.next; insertPosition.next = firstLarger; insertPosition = insertPosition.next; firstLarger = current.next; } } } newHead = newHead.next; return
newHead; }} |
leetcode--Partition List,布布扣,bubuko.com
原文地址：http://www.cnblogs.com/averillzheng/p/3779024.html
