[Leetcode] 475. Heaters 解题报告

最新推荐文章于 2021-01-01 11:50:27 发布

魔豆Magicbean

最新推荐文章于 2021-01-01 11:50:27 发布

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分类专栏： IT公司面试习题文章标签： Leetcode 解题报告

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本文介绍了一种算法，用于计算使所有房屋都能被覆盖的最小加热半径。通过排序房屋和加热器位置，并找到每个房屋最近的加热器来确定半径。

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题目：

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.

思路：

首先对house和heater进行排序（这样在后面可以直接用线性扫描）。对于每个house，我们找距离它最近的heater，然后更新radius即可。由于house的位置顺序是递增的，所以最近的heater的位置也一定是递增的。算法的时间复杂度是O(mlogm + nlogn)，其中m和n分别是house和heater的数量。

代码：

class Solution {
public:
    int findRadius(vector<int>& houses, vector<int>& heaters) {
        sort(houses.begin(), houses.end());
        sort(heaters.begin(), heaters.end());
        int heater_pos = 0, ans = 0;
        for (auto h : houses) {
            while (heater_pos + 1 < heaters.size() && abs(heaters[heater_pos + 1] - h) <= abs(heaters[heater_pos] - h)) {
                ++heater_pos;           // find the nearest heater
            }
            ans = max(ans, abs(heaters[heater_pos] - h));
        }
        return ans;
    }
};