[LeetCode] Remove Duplicates from Sorted Array II 解题报告

Follow up for “Remove Duplicates”:
What if duplicates are allowed at most twice[/i]?
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]

.
» Solve this problem

[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<n)
8:      {
9:        if(A[cur] == A[cur-1])
10:        {
11:          if(occur >=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:       }