public int GetNumberOfK(int [] array , int k) {
        int left = 0;
        int right = array.length - 1;

        //1. 使用二分法确定左边界,即第一个等于k的索引位置
        while(left < right){
            int mid = (right - left)/2 + left;
            if(array[mid] >= k){
                right = mid;
            }else{
                left = mid + 1;
            }
        }

        //2. 遍历统计等于k的元素个数
        int ans = 0;
        while(left < array.length && array[left] == k){
            ans++;
            left++;
        }

        return ans;
    }