给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

进阶：

你可以设计并实现时间复杂度为 O(log n) 的算法解决此问题吗？

示例 1：

输入：nums = [5,7,7,8,8,10], target = 8
输出：[3,4]

示例 2：

输入：nums = [5,7,7,8,8,10], target = 6
输出：[-1,-1]

示例 3：

输入：nums = [], target = 0
输出：[-1,-1]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array

python

class Solution:
    def searchRange(self, nums: [int], target: int) -> [int]:

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

        # 找到第一个=target的位置
        leftIdx = search(nums, target)
        # 找到第一个 > target的位置
        rightIdx = search(nums, target+1)
        if leftIdx == len(nums) or nums[leftIdx] != target:
            return [-1, -1]
        return [leftIdx, rightIdx-1]

if __name__ == "__main__":
    nums1 = [1,3]
    target1 = 2
    nums2 = [1,3,3,3,3,3,3,3,5,5,6]
    target2 = 5
    test = Solution()
    print(test.searchRange(nums1,target1)) # [-1, -1]
    print(test.searchRange(nums2,target2)) # [8, 9]

golang

func main() {
	var nums = []int{1, 3, 3, 3, 3, 3, 3, 3, 5, 5, 6}
	var target int = 5
	res := searchRange(nums, target)
	fmt.Println(res)
}

func searchRange(nums []int, target int) []int {
	//第一个=target的位置
	leftIdx := search(nums, target)
	rightIdx := search(nums, target+1)

	if leftIdx == len(nums) || nums[leftIdx] != target {
		return []int{-1, -1}
	}
	return []int{leftIdx, rightIdx - 1}
}

func search(nums []int, target int) int {
	var left int = 0
	var right int = len(nums)
	for left < right {
		var mid int = (left + right) / 2
		if nums[mid] >= target {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}