[leetcode]Partition List

Partition List

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.

算法思路：

维护两个指针，一个指向带插入位置的前驱，一个顺序遍历list，遇到>=x的节点，next，遇到<x的就把它插入到上一个指针的后面。第一个节点<x时候，需要特殊处理

public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null || head.next == null)    return head;
        ListNode hhead = new ListNode(0);
        hhead.next = head;
        ListNode pre = hhead;
        ListNode tail = hhead;
        while (tail.next != null) {
            if (tail.next.val < x) {//第一个节点特殊处理
                if(tail == hhead){
                    pre = pre.next;
                    tail = tail.next;
                    continue;
                }
                ListNode tem = tail.next;
                tail.next = tem.next;
                tem.next = pre.next;
                pre.next = tem;
                pre = pre.next;
                if(tail.val < x){
                    tail = tail.next;
                }
            } else
                tail = tail.next;
        }
        return hhead.next;
    }
}

感觉这道题做的很垃圾。。。。