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86. Partition List

时间:2017-10-25 11:40:55      阅读:105      评论:0      收藏:0      [点我收藏+]

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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.

 

 1 class Solution {
 2     public ListNode partition(ListNode head, int x) {
 3         if(head==null||head.next==null) return head;
 4         ListNode big = new ListNode(0);
 5         ListNode small = new ListNode(0);
 6         ListNode smallhead = small;
 7         ListNode bighead = big;
 8         while(head!=null){
 9             if(head.val<x){ 
10                 small.next = head;
11                 small = head;
12             }
13             else{
14                 big.next = head;
15                 big = head;
16             }
17             head = head.next;
18         }
19         big.next = null;
20         small.next = bighead.next;
21         return smallhead.next;
22     }
23    
24 }

 

86. Partition List

标签:less   class   content   script   head   blog   you   public   serve   

原文地址:http://www.cnblogs.com/zle1992/p/7727216.html

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