LeetCode 66 Partition List

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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;    }}

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