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].

这道题跟Remove Duplicates from Sorted Array比较类似,区别只是这里元素可以重复出现至多两次,而不是一次。其实也比较简单,只需要维护一个counter,当counter是2时,就直接跳过即可,否则说明元素出现次数没有超,继续放入结果数组,若遇到新元素则重置counter。总体算法只需要扫描一次数组,所以时间上是O(n),空间上只需要维护一个index和counter,所以是O(1)。

  1. class Solution {
  2. public:
  3. int removeDuplicates(int A[], int n) {
  4. if(A==NULL||n<) return ;
  5. int count=;
  6. int res=;
  7. for(int i=;i<n;i++){
  8. if(i>&&A[i]==A[i-]){
  9. count++;
  10. }else
  11. count=;
  12. if(count<){
  13. A[res++]=A[i];
  14. }
  15. }
  16. return res;
  17. }
  18. };

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