How to prevent timeouts--remove duplicates from Sorted Array i&&ii

Given a sorted array, remove the duplicates in place such, all element appear only once and return the new L Ength.

Do the allocate extra space for another array, and you must does this on place with constant memory.

For example,
Given input array nums = [1,1,2] ,

Your function should return length = 2 , with the first of the elements of nums being and 1 2 Respectivel Y. It doesn ' t matter what are you leave beyond the new length.

Class Solution {public:    int removeduplicates (int a[], int n) {       if (!n) return NULL;          int num=1,i;          for (I=1;i<n;++i)              if (a[i]!=a[i-1])                  a[num++]=a[i];            return num;              }};

II. Add a Count variable to record the number of occurrences of key.

Class Solution {public:     int removeduplicates (int a[], int n) {          //Start Typing your C + + solution below          //Do Not write int main () function          if (n = = 0)              return 0;                        int key = a[0];         int count = 0;         int start = 0;                  for (int i = 0; i < n; i++)             if (key = = A[i])                 count++;             else             {                 for (int j = 0; J < min (2, count); j + +)                     a[start++] = key;                 key = A[i];                 Count = 1;             }                      for (int j = 0; J < min (2, count); j + +)             a[start++] = key;                    return start;     } };

　　

How to prevent timeouts--remove duplicates from Sorted Array i&&ii