Partition-List.md

一、题目

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.

给定一个单链表和数值x，划分链表使得所有小于x的节点排在大于等于x的节点之前。

你应该保留两部分内链表节点原有的相对顺序。

二、解题思路

依据题意，是要根据值x对链表进行分割操作，具体是指将所有小于x的节点放到不小于x的节点之前，咋一看和快速排序的分割有些类似，但是这个题的不同之处在于只要求将小于x的节点放到前面，而并不要求对元素进行排序。

这种分割的题使用两路指针即可轻松解决。左边指针指向小于x的节点，右边指针指向不小于x的节点。由于左右头节点不确定，我们可以使用两个dummy节点。

三、解题代码

/**
 * 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) {
        ListNode leftDummy = new ListNode(0);
        ListNode leftCurr = leftDummy;
        ListNode rightDummy = new ListNode(0);
        ListNode rightCurr = rightDummy;

        ListNode runner = head;
        while (runner != null) {
            if (runner.val < x) {
                leftCurr.next = runner;
                leftCurr = leftCurr.next;
            } else {
                rightCurr.next = runner;
                rightCurr = rightCurr.next;
            }
            runner = runner.next;
        }

        // cut off ListNode after rightCurr to avoid cylic
        rightCurr.next = null;
        leftCurr.next = rightDummy.next;

        return leftDummy.next;
    }
}