leetcode -- 86. Partition List

题目描述

题目难度:Medium
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.

• Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

AC代码

用两个链表分别保存小于x的序列和大于等于x的序列

/**
 * 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 less = null; //小于x的链表
        ListNode more = null; // 大于x的链表
        ListNode moreHead = null;
        ListNode lessHead = head;
        while(head != null){ 
            if(head.val < x){
                if(less == null) {
                    less = head;
                    lessHead = less;
                }
                else{
                    less.next = head;
                    less = less.next;
                } 
            }
        else{
            if(more == null) {
                more = head;
                moreHead = more;
            }
            else{
                more.next = head;
                more = more.next;
            }
        }
            head = head.next;
        }
        if(more != null) more.next = null //很关键,少了可能会造成无限循环
        if(less != null) less.next = moreHead;
        return lessHead;
    }
}