【题目描述】
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,
Given1->4->3->2->5->2and x = 3,
return1->2->2->4->3->5.
【解析】
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。
【代码】
package leetcode.easy.easy;public class Partition_List {public class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}public static void main(String[] args) {}public ListNode partition(ListNode head, int x) {if(head==null||head.next==null)return head;ListNode small = new ListNode(-1);ListNode newsmallhead = small;ListNode big = new ListNode(-1);ListNode newbighead = big;while(head!=null){if(head.val<x){small.next = head;small = small.next;}else{big.next = head;big = big.next;}head = head.next;}big.next = null;small.next = newbighead.next;return newsmallhead.next;}
}