Follow up for "Remove Duplicates":

What if duplicates are allowed at most 

twice?

For example,

Given sorted array A = 

[1,1,1,2,2,3],

Your function should return length = 

5, and A is now 

[1,1,2,2,3].

[解题思路]

加一个变量track一下字符出现次数即可，这题因为是已经排序的数组，所以一个变量即可解决。但是如果是没有排序的数组，可以引入一个hashmap来处理出现次数。

[Code]

1:    int removeDuplicates(int A[], int n) {  2:      // Start typing your C/C++ solution below  3:      // DO NOT write int main() function  4:      if(n<=1) return n;  5:      int pre=1, cur =1;  6:      int occur = 1;  7:      while(cur
    
     =2)  12:          {  13:            cur++;          14:            continue;  15:          }  16:          else  17:          {  18:            occur++;  19:          }  20:        }  21:        else  22:        {  23:          occur = 1;   24:        }        25:        A[pre] = A[cur];  26:        pre++;  27:        cur++;        28:      }  29:      return pre;  30:    }
    

Update 03/09/2014  improve readability a bit.

1:       int removeDuplicates(int A[], int n) {  2:            if(n == 0) return 0;  3:            int occur = 1;  4:            int index = 0;  5:            for(int i =1; i< n; i++)  6:            {  7:                 if(A[index] == A[i])  8:                 {  9:                      if(occur == 2)  10:                      {  11:                           continue;  12:                      }  13:                      occur++;  14:                 }  15:                 else  16:                 {  17:                      occur =1 ;  18:                 }  19:                 A[++index] = A[i];  20:            }  21:            return index+1;  22:       }