leetcode [34] 在排序数组中查找元素的第一个和最后一个位置

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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。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

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]

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        if len(nums) == 0:
            return [-1, -1]

        low, high = 0, len(nums)
        while low < high:
            mid = (low + high) // 2
            if nums[mid] == target:
                low = high = mid
                while low - 1 >= 0 and nums[low - 1] == target:
                    low -= 1
                while high + 1 < len(nums) and nums[high + 1] == target:
                    high += 1
                return [low, high]
            if nums[mid] < target:
                low = mid + 1
            else:
                high = mid
        return [-1, -1]

优化：

由于找到相等的时候，退化为线性查找，优化为对数查找， 注意细节：

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        if len(nums) == 0:
            return [-1, -1]

        low, high = 0, len(nums)
        while low < high:
            mid = (low + high) // 2
            if nums[mid] == target:
                result = []
                left, right = low, mid
                # 查找左边界
                while left < right:
                    newmid = (left + right) // 2
                    if nums[newmid] < target:
                        left = newmid + 1
                    else:
                        right = newmid

                result.append(left)
                left, right = low, high
                # 查找右边界
                while left < right:
                    newmid = (left + right) // 2
                    if nums[newmid] > target:
                        right = newmid
                    else:
                        left = newmid + 1
                result.append(right - 1)
                return result
            if nums[mid] < target:
                low = mid + 1
            else:
                high = mid