LeetCode 66 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.

分析：

这就是快速排序的思路，不过是在链表上进行，维护一个tail指针，之前放的都是小于x的节点，往后遍历一遍，碰见>=x的跳过，碰见<x的移到tail后面，再把tail后移一位。

/**
* 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 newHead = new ListNode(0);
newHead.next = head;
ListNode tail = newHead;
ListNode p = newHead;
while(p.next != null){
if(p.next.val < x){
if(p==tail){//no need to move
tail = tail.next;
p = p.next;
}else{
ListNode tmp = p.next;
p.next = p.next.next;//delete
tmp.next = tail.next;//insert after tail
tail.next = tmp;
tail = tail.next;
}

}else{
p = p.next;
}
}
return newHead.next;
}
}