O(lgn)时间内找出有序数组中某个元素出现的次数

题目：

      找出有序数组中指定元素出现的次数，要求时间复杂度为O(lgn)

      ex，

      数组｛0,0,0,2,3,3,3,3,3,4,5,5｝，0出现3次，3出现5次

 

思路：

很容易想到的一个办法是binary_search找到指定元素，然后左右查询，得到出现的次数k，但其时间复杂度为O(lgn)+k。

      可通过改进binary_search，做两次查找，一次得到指定元素的起始出现位置，一次得到终止出现位置。

 

usingnamespace std;

int binary_search_my(int arr[],int p,int q,int value,bool firstflag =true)
{
int begin = p;
int end = q;
while(begin <= end)
{
int mid = (begin + end)/2;
if(arr[mid] == value)
{
if(firstflag)
{
if(mid != p && arr[mid-1] != value)
return mid;
elseif(mid == p)
return p;
else
end = mid - 1;
}
else
{
if(mid != q && arr[mid+1] != value)
return mid;
elseif(mid == q)
return q;
else
begin = mid + 1;
}
}
elseif(arr[mid] < value)
begin = mid + 1;
else
end = mid - 1;
}
return -1;
}

int CountNumberOfOccurancesInSortedArr(int arr[],int n,int value)
{
int last = binary_search_my(arr, 0, n-1, value,false);
int first = binary_search_my(arr, 0, n-1, value,true);
if(last == -1 && first == -1)
return 0;
else
return last - first + 1;
}

int main()
{
int arr[] = {0,0,0,2,3,3,3,3,3,4,5,5};
int nsize =sizeof(arr)/sizeof(int);
copy(arr, arr+nsize, ostream_iterator<int>(cout,"/t"));
cout<<endl;
cout<<"Count number of 0: "<<(CountNumberOfOccurancesInSortedArr(arr, nsize, 0))<<endl;
cout<<"Count number of 3: "<<(CountNumberOfOccurancesInSortedArr(arr, nsize, 3))<<endl;
cout<<"Count number of 5: "<<(CountNumberOfOccurancesInSortedArr(arr, nsize, 5))<<endl;
cout<<"Count number of 2: "<<(CountNumberOfOccurancesInSortedArr(arr, nsize, 2))<<endl;
cout<<"Count number of 1: "<<(CountNumberOfOccurancesInSortedArr(arr, nsize, 1))<<endl;
return 0;
}