leetcode : Partition List

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.

思路: 两个指针

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head == null || head.next == null) {
            return head;
        }
        
        ListNode great = head;
        ListNode less = head;
        
        ListNode dummy1 = new ListNode(0);
        dummy1.next = less;
        ListNode dummy2 = new ListNode(0);
        dummy2.next = great;
        
        while(head != null) {
            if(head.val < x) {
                less.next = head;
                less = less.next;
            } else {
                great.next = head;
                great = great.next;
            }
            head = head.next;
        }
        
        less.next = dummy2.next;
        
        return dummy1.next;
    }
}

  

posted @ 2017-03-19 21:37  notesbuddy  阅读(132)  评论(0编辑  收藏  举报