leetcode -- Search for a Range

本文介绍了如何通过二分查找算法在有序数组中高效地找到给定目标值的起始和结束位置。对于未找到目标值的情况,返回[-1, -1]。详细步骤包括递归出口条件以及处理目标值大于中间元素的情况。

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Q: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].

分析:因为是有顺序序列,应该可以通过二分法查找.两个递归出口:(1)n=1且没有目标找到;(2)已经找到目标.
需要注意的是target>A[n/2]的情况,这时对[A+n/2, n-n/2]子序列重新递归查找,返回值target位置要加上之前的n/2.

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ret;
        if(n == 1 && target !=A[0])   //递归出口1,找不到目标值
            ret.assign(2, -1);
        else if(target == A[n/2]){      //递归出口2,找到目标值
            int tbeg = n/2, tend = n/2;
            while((--tbeg) >= 0 && target == A[tbeg]);
            while((++tend) < n && target == A[tend]);
            ret.push_back(tbeg + 1);
            ret.push_back(tend - 1);
        }
        else if(target < A[n/2])
            ret = searchRange(A, n/2, target);
        else{
	    ret = searchRange(A + n/2, n - n/2, target);
	    if(ret[0] == -1 && ret[1] == -1)  return ret;
	    ret[0] += n/2; ret[1] += n/2;       //***attention***
        }
        return ret;
    }
};


### 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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