题解 | #数字在升序数组中出现的次数#

class Solution {
  private:
    int findIndexNums(vector<int>& nums, int index) {
        /*
            找到和对应下标的值相同的元素个数
        */
       // 向左寻找
       int leftIndex = index;
       int rightIndex = index;
       while(leftIndex>=0 && nums[leftIndex] == nums[index]){  // 向左寻找 
            leftIndex--;
       }
       while (rightIndex<=nums.size()-1 && nums[rightIndex] == nums[index]){  // 向右寻找
            rightIndex++;
       }
       return rightIndex-leftIndex-1;

    }
  public:
    int GetNumberOfK(vector<int>& nums, int k) {
        // 数组为空 或 k不在nums中  均返回0
        if (nums.empty()) return 0;

        int left = 0;
        int right = nums.size() - 1;
        int midIndex;

        while (left <= right) {

            midIndex = left + (right - left) / 2;

            if (nums[midIndex] == k) {
                int result = findIndexNums(nums, midIndex);  // 计算同一值出现的次数
                return result;
            } else if (nums[midIndex] > k) {
                // left = midIndex+1;
                right = midIndex-1;
            } else if (nums[midIndex] < k) {
                // right = midIndex-1;
                left = midIndex+1;
            }
        }
        return 0;
    }
};