Partition List - Leetcode

/**
* 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 left=new ListNode(-1), right=new ListNode(-1);
ListNode cur = head, left_head = null, right_head = null;
while(cur!=null){
if(cur.val < x){
left.next = cur;
cur = cur.next;
left = left.next;
left.next = null;
if(left_head == null)
left_head = left;
}else{
right.next = cur;
cur = cur.next;
right = right.next;
right.next = null;
if(right_head == null)
right_head=right;
}
}

if(right_head!=null){
left.next = right_head;
left = left.next;
if(left_head == null) //只有右边
left_head = left;
}
return left_head;
}
}

分析：就一个链表分区，

第一个想到，分别从链表两端看往中间走，左边指针放比pivot小的，右边指针放比pivot大的。

1）如果左边遇到大的，右边遇到小的：左右调换；

2）如果左边遇到大的，右边遇到大的：左边插入右边；

3）如果左边遇到小的，右边遇到小的：右边插入左边；

直到左指针遇到右指针。

因为是链表结构，交换/插入的实现都是很繁琐。舍弃！

第二个想法是链表从左往右走，当遇到小的放到左指针后面，当遇到大的放到右指针后面。最后将左指针连接右指针。

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

.