LeetCode#86. Partition List

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题目：链表分块（比target小的在左边，比target大的在右边，并且元素之间保持原来的顺序）
难度：Medium
思路：定义两个头结点，分别将比target小、不小于target的节点连接，遍历完原始链表后需要释放left后面的元素
代码：

/** * 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 || head.next == null){            return head;        }        ListNode leftHead = new ListNode(0);        ListNode rightHead = new ListNode(0);        ListNode left = leftHead;        ListNode right = rightHead;        ListNode node = head;        while(node != null){            if(node.val < x){                left.next = node;                left = left.next;            }else{                right.next = node;                right = right.next;            }            node = node.next;;        }        left.next = rightHead.next;        right.next = null;        return leftHead.next;    }}

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