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 1, 1, 2, 2 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 }