https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/
Medium
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
- 二分查找，需要理解对左右index上下界限如何设置。需要对左右index各做一次二分查找。
- 查找左index时，如果遇到nums[middle] == target的情况，则还需要对左半边继续查找，因为要找到target第一次出现位置。
- 查找右index时，选择查找target右边的第一个index，即target最后一次出现位置+1，所以在主程序里对返回结果-1。
- https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/
1 class Solution: 2 def searchRange(self, nums: List[int], target: int) -> List[int]: 3 if nums is None: 4 return [-1, -1] 5 6 def helper(nums, target, search_left_index): 7 left, right = 0, len(nums) 8 9 while left < right: 10 middle = (left + right) // 2 11 12 # check if searching the leftmost index or not 13 if nums[middle] > target or (search_left_index and nums[middle] == target): 14 right = middle 15 else: 16 left = middle + 1 17 18 return left 19 20 left_index = helper(nums, target, True) 21 22 if left_index == len(nums) or nums[left_index] != target: 23 return [-1, -1] 24 25 return [ left_index, helper( nums, target, False ) - 1 ]