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;elsereturn 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;}