LeetCode 86. Partition List

原题链接在这里：https://leetcode.com/problems/partition-list/

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.

题解：

Two pointers, 若是当前点val比x小，就连在连到left dummy head后面。否则连到right dummy head后面.

Time Complexity: O(n). Space: O(1).

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode partition(ListNode head, int x) {
11         if(head == null || head.next == null){
12             return head;
13         }
14         
15         ListNode leftDummy = new ListNode(0);   //左侧dummy后面链接的点val比x小
16         ListNode rightDummy = new ListNode(0);  //右侧dummy后面链接的店val比x大或者相等
17         
18         ListNode leftCur = leftDummy;
19         ListNode rightCur = rightDummy;
20         while(head != null){
21             if(head.val < x){
22                 leftCur.next = head;
23                 leftCur = leftCur.next;
24                 head = head.next;
25             }else{
26                 rightCur.next = head;
27                 rightCur = rightCur.next;
28                 head = head.next;
29             }
30         }
31         
32         rightCur.next = null;   //断开右侧尾节点
33         leftCur.next = rightDummy.next;
34         return leftDummy.next;
35     }
36 }