[Leetcode] Partition List

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.

Solution:

1. 从左往右扫描，找到第一个大于X的指针，然后再该指针左边，不断插入小于X的元素。这里为了避免处理head是否为空的检测，在头指针位置先插入一个干扰元素，以保证head永不为空，然后在最后返回的时候删除掉。
2. 用small来指向比x小的值的最后一个，用big来指向大于或等于x的值的最后一个，即下一个应该插入的位置。

/**
 * 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 root=new ListNode(-1);
        root.next=head;
        ListNode small=root;
        ListNode big=root;
        ListNode pointer=root;
        while(pointer.next!=null){                     //在这里找第一个比x大或等的数
            if(pointer.next.val>=x){
                small=pointer;
                big=pointer.next;
                pointer=pointer.next;
                break;
            }
            pointer=pointer.next;
        }
        while(pointer.next!=null){
            if(pointer.next.val>=x){
                pointer=pointer.next;
                big=big.next;
            }else{
                pointer=pointer.next;
                ListNode temp=pointer.next;
                pointer.next=small.next;
                small.next=pointer;
                big.next=temp;
                pointer=big;
                small=small.next;
            }
        }
        return root.next;
    }
}