题目：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]。
链接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
法一：自己的代码
思路：虽然right=len(nums)-1，即左闭右闭区间，但没有用left<right终止循环，当left=mid时，此时就做出判断终止循环，即可以灵活应用，不是必须当left==right的时候终止循环，
from typing import List class Solution: def searchRange(self, nums: List[int], target: int) -> List[int]: right = len(nums) - 1 if right == -1: return [-1, -1] elif right == 0: if nums[0] == target: return [0,0] else: return [-1, -1] # 先查找左端点 def find_left(left, right): while left < right: mid = (left+right) >> 1 # 等于的时候，目标值如果存在一定在等于的这两个值里 if left == mid: if nums[right] == target: if nums[left] == target: return left else: return right elif nums[left] == target: return left else: return -1 if nums[mid] >= target: right = mid else: left = mid + 1 return left if nums[left] == target else -1 # return find_left(0, right) # 再查找右端点 def find_right(left, right): while left < right: mid = (left+right) >> 1 if left == mid: if nums[left] == target: if nums[right] == target: return right else: return left elif nums[right] == target: return right else: return -1 # 只有大于才让右端点左移， if nums[mid] > target: right = mid else: left = mid k = find_left(0, right) if k == -1: return [-1, -1] else: return [k, find_right(0, right)]
改进后的代码，right=len(nums)，左闭右开区间
from typing import List class Solution: def searchRange(self, nums: List[int], target: int) -> List[int]: right = len(nums) l = right # 找左侧边界的二分查找 def find_left(left, right): # 搜索区间最短的时候有一个元素，为左边的left while left < right: mid = (left + right) >> 1 if nums[mid] == target: right = mid # 右边是开区间，而mid一定不满足条件，所以right直接取mid elif nums[mid] > target: right = mid # 左边是闭区间，mid一定不满足条件，直接将其排除，left取mid+1 # 综上可见，left和right本质上都一样，都是对已经作过判断的位置mid # 不再进行判断，但是由于left闭right开，导致了排序mid位置的方法不一样 elif nums[mid] < target: left = mid + 1 if left == l: return -1 return left if nums[left] == target else -1 # 找右侧边界的二分查找 def find_right(left, right): while left < right: mid = (left + right) >> 1 # 注意这里要加1，因为left是闭区间 # 如果不加1会出现死循环，如left等于2，right等于3时，这时mid每次都是left，陷入死循环 if nums[mid] == target: left = mid + 1 # 如果大于目标值，则该值一定不满足条件， elif nums[mid] > target: right = mid # 如果小于，右移一个，因为左边是闭区间， elif nums[mid] < target: left = mid + 1 # 注意当left == right的时候，循环结束，但由于nums[mid] > target时，right=mid，即nums[right]一定不等于target # 也就是说nums[left-1]才可能等于target，故需要判断nums[left-1] if nums[max(left-1, 0)] == target: return max(left-1, 0) else: return -1 k = find_left(0, right) if k == -1: return [-1,-1] else: return [k, find_right(0, right)] if __name__ == '__main__': solution = Solution() # result = solution.searchRange(nums = [5,7,7,8,8,8,8,9,10], target = 8) # result = solution.searchRange(nums = [7,7,7,7], target = 7) result = solution.searchRange(nums = [-4, -2, -1], target = -5) # result = solution.searchRange(nums = [5,7,7,8,8,10], target = 8) # result = solution.searchRange(nums = [1,2,3,4,5,8,8,8,8,8,8,8], target = 8) # result = solution.searchRange(nums = [1,1,1,1,1,1,1,1,1,2,3], target = -22) # result = solution.searchRange(nums = [0], target = 0) # result = solution.searchRange(nums = [8,8,8,8,8,8,8], target = 8) print(result)
ttt