LeetCode #475 - Heaters

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本文介绍了一种算法，用于计算一组加热器覆盖所有房屋所需的最小供暖半径。通过排序和搜索，确保每个房屋都能被至少一个加热器覆盖。

题目描述：

Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.

Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.

So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.

Note:

Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
As long as a house is in the heaters' warm radius range, it can be warmed.
All the heaters follow your radius standard and the warm radius will the same.

Example 1:

Input: [1,2,3],[2]
Output: 1
Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.

Example 2:

Input: [1,2,3,4],[1,4]
Output: 1
Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.

给定一组加热器的位置和一组房子的位置，求加热器的供暖半径使得所有的房子都能取暖。先将加热器和房子都排序，那么对于每栋房子都在加热器数组中搜索，找到距离最近的加热器，这就是为了给这栋房子供暖所需要的最小的供暖半径，对每栋房子都这样搜索就可以确定满足所有房子的供暖半径。

class Solution {
public:
    int findRadius(vector<int>& houses, vector<int>& heaters) {
        if(houses.size()==0||heaters.size()==0) return 0;
        sort(houses.begin(),houses.end());
        sort(heaters.begin(),heaters.end());
        int result=0;
        for(int i=0;i<houses.size();i++)
        {
            int j=lower_bound(heaters.begin(),heaters.end(),houses[i])-heaters.begin();
            if(j==0) result=max(result,heaters[0]-houses[i]);
            else if(j==heaters.size()) result=max(result,houses[i]-heaters.back());
            else result=max(result,min(heaters[j]-houses[i],houses[i]-heaters[j-1]));
        }
        return result;
    }
};