【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.

题意：给定一个单链表和一个x，把链表中小于x的放到前面，大于等于x的放到后面，每部分元素的原始相对位置不变。

思路：其实很简单，遍历一遍链表，把小于x的都挂到head1后，把大于等于x的都放到head2后，最后再把大于等于的链表挂到小于链表的后面就可以了。

这道题不难，主要是给不熟悉指针的同学学习交流。

/**
 * 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) {
        ListNode lessHead = new ListNode(0);
        ListNode greaterHead = new ListNode(0);
        ListNode node = head, less = lessHead, greater = greaterHead;
        while (node != null) {
            ListNode next = node.next;
            if (node.val < x) {
                less.next = node;
                less = less.next;
                less.next = null;
            } else {
                greater.next = node;
                greater = greater.next;
                greater.next = null;
            }
            node = next;
        }
        less.next = greaterHead.next;
        return lessHead.next;
    }
}

【LeetCode】Partition List 解题报告