Problem 86: Partition List

https://leetcode.com/problems/partition-list/

类比: Partion Array 这道题

http://www.lintcode.com/en/problem/partition-array/

思路

• 创建一个left LinkedList，一个right LinkedList。

• 遇到小的挂left上，遇到大的挂right上

• 最后合并 left 和 right

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) {
            return head;
        }

        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy;
        ListNode right = rightDummy;

        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = head;
            } else {
                right.next = head;
                right = head;
            }
            head = head.next;
        }

        right.next = null;
        left.next = rightDummy.next;

        return leftDummy.next;
    }
}

易错点

1. 两个指针，left用来探路，leftDummy用来标记“头”

   ListNode left = leftDummy;
   ListNode right = rightDummy;

2. LinkedList的循环

   while (head != null) {
    执行语句;
    head = head.next;
   }

   其中，head = head.next;作为增加变量使用的。

3. 掐头去尾进行拼接

   right.next = null;
   left.next = rightDummy.next;