[LeetCode]--34. Search for a Range

在给定的有序整数数组中,利用二分查找算法找到指定目标值的起始和结束位置。算法的时间复杂度需为O(log n)。如果目标值未在数组中,则返回[-1, -1]。例如,通过讨论区学习到的简洁解法。" 54782461,5762545,SQL递归查询实战:从概念到路径构造,"['SQL查询', '数据库操作', '递归']

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Given a sorted array of integers, 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].

For example,
Given

[5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
public int[] searchRange(int[] nums, int target) {
        int low = 0, high = nums.length - 1, temp = -1;
   
### LeetCode 475 Heaters Problem Solution and Explanation In this problem, one needs to find the minimum radius of heaters so that all houses can be warmed. Given positions of `houses` and `heaters`, both represented as integer arrays, the task is to determine the smallest maximum distance from any house to its nearest heater[^1]. To solve this issue efficiently: #### Binary Search Approach A binary search on answer approach works well here because increasing the radius monotonically increases the number of covered houses. Start by sorting the list of heaters' locations which allows using binary search for finding closest heater distances quickly. ```python def findRadius(houses, heaters): import bisect houses.sort() heaters.sort() max_distance = 0 for house in houses: pos = bisect.bisect_left(heaters, house) dist_to_right_heater = abs(heaters[pos] - house) if pos < len(heaters) else float('inf') dist_to_left_heater = abs(heaters[pos-1] - house) if pos > 0 else float('inf') min_dist_for_house = min(dist_to_right_heater, dist_to_left_heater) max_distance = max(max_distance, min_dist_for_house) return max_distance ``` This code snippet sorts the lists of houses and heaters first. For each house, it finds the nearest heater either directly or indirectly (to the left side). It calculates the minimal distance between these two options and updates the global maximal value accordingly[^3]. The constraints specify that numbers of houses and heaters do not exceed 25000 while their positions range up to \(10^9\)[^2], making efficient algorithms like binary search necessary due to large input sizes involved.
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