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首先是二分查找

java版本

class Solution {
       public int search(int[] nums, int target) {
           int pivot, left = 0, right = nums.length - 1;        while (left <= right) {
               pivot = left + (right - left) / 2;            if (nums[pivot] == target) return pivot;            if (target < nums[pivot]) right = pivot - 1;            else left = pivot + 1;        }        return -1;    }}

python版本

class Solution:    def search(self, nums: List[int], target: int) -> int:        left, right = 0, len(nums) - 1        while left <= right:            pivot = left + (right - left) // 2            if nums[pivot] == target:                return pivot            if target < nums[pivot]:                right = pivot - 1            else:                left = pivot + 1        return -1

判断一个有序数组中一个数字出现的次数

参考链接：https://blog.csdn.net/nwpubear/article/details/81915247