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:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode leftDummy = new ListNode(0);   //左侧dummy后面链接的点val比x小
        ListNode rightDummy = new ListNode(0);  //右侧dummy后面链接的店val比x大或者相等
        
        ListNode leftCur = leftDummy;
        ListNode rightCur = rightDummy;
        while(head != null){
            if(head.val < x){
                leftCur.next = head;
                leftCur = leftCur.next;
                head = head.next;
            }else{
                rightCur.next = head;
                rightCur = rightCur.next;
                head = head.next;
            }
        }
        
        rightCur.next = null;   //断开右侧尾节点
        leftCur.next = rightDummy.next;
        return leftDummy.next;
    }
}