【原创】leetCodeOj ---Partition List 解题报告

原题地址：

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

 

题目内容：

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.

 

方法：

链表的基本功。

维护两个带头结点的链表，一个存 < ，一个存 >=，然后一个连接就可以了。

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null)
            return null;
        ListNode lessHead = new ListNode(0);
        ListNode moreHead = new ListNode(0);
        ListNode lessTail = lessHead;
        ListNode moreTail = moreHead;
        ListNode tmp = null;
        while (head != null)
        {
            tmp = head;
            head = head.next;
            tmp.next = null;            
            if (tmp.val < x)
            {

                lessTail.next = tmp;
                lessTail = lessTail.next;
            }
            else
            {
                moreTail.next = tmp;
                moreTail = moreTail.next;
            }
        }
        if (lessHead.next != null)
            lessTail.next = moreHead.next;
        else
            lessHead.next = moreHead.next;
        return lessHead.next;
    }
}