LeetCode 80. Remove Duplicates from Sorted Array II (从有序序列里移除重复项之二)

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1122 and 3. It doesn't matter what you leave beyond the new length.

 

 


题目标签:Array
  这道题目和之前的区别就是,可以保留第二个重复的数字。基本想法都和前一题一样,在这里只要多加一个int count 来记录这是第几个重复的number,如果是第二个的话,把pointer 往右移动一格,并且把nums[i] 的值 复制到 nums[pointer],然后count++。除此之外,还需要在遇到不同数字的情况里,加上,count = 1, 因为一旦遇到不同的数字,那么count 计数又要重新开始了。详细可以看代码,和之前那题代码的比较。
 
 
 

Java Solution:

Runtime beats 27.80% 

完成日期:07/29/2017

关键词:Array

关键点:多设一个int count 来记录出现重复数字的次数

 

 1 public class Solution 
 2 {
 3     public int removeDuplicates(int[] nums) 
 4     {
 5         if(nums.length <= 2)
 6             return nums.length;
 7         
 8         int pointer = 0;
 9         int count = 1;
10         
11         for(int i=1; i<nums.length; i++)
12         {
13             // if this number is different than pointer number
14             if(nums[i] != nums[pointer])
15             {
16                 pointer++;
17                 nums[pointer] = nums[i];
18                 count = 1;
19             }
20             else // if this number is same as pointer number
21             {
22                 if(count == 1) // if it is second same number
23                 {
24                     pointer++;
25                     nums[pointer] = nums[i];
26                     count++;
27                 }
28             }
29         }
30         
31         return pointer + 1;
32     }
33 }

参考资料:N/A

 

LeetCode 算法题目列表 - LeetCode Algorithms Questions List

 

posted @ 2017-07-30 05:33  Jimmy_Cheng  阅读(155)  评论(0编辑  收藏  举报