LintCode 1219: Heaters

最新推荐文章于 2025-12-22 09:47:08 发布

原创最新推荐文章于 2025-12-22 09:47:08 发布 · 235 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法 #leetcode

algorithm-design 专栏收录该内容

817 篇文章

订阅专栏

本文探讨了一种算法解决方案，旨在确定一组固定位置加热器的最小辐射范围，以确保所有房屋都能得到供暖。通过分析房屋与加热器的位置，采用二分查找和比较最小距离的方法，有效地解决了这一问题。

1219. Heaters

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.

Example

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.

Notice

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

Input test data (one parameter per line)How to understand a testcase?

解法1：
我最开始的做法是用二分查找，找到最合适的那个target值。

class Solution {
public:
    /**
     * @param houses: positions of houses
     * @param heaters: positions of heaters
     * @return: the minimum radius standard of heaters
     */
    int findRadius(vector<int> &houses, vector<int> &heaters) {
       int num_house = houses.size();
       int num_heater = heaters.size();
       int radius = 0;
       sort(heaters.begin(), heaters.end());
       int start = 0, end = max(abs(heaters[num_heater - 1] - houses[0]), abs(heaters[0] - houses[num_house - 1]));

       while(start + 1 < end) {
           int mid = start + (end - start) / 2;
           if (satisfied(houses, heaters, mid)) {
               end = mid;
           } else {
               start = mid;
           }
       }
       
       if (satisfied(houses, heaters, start)) return start;
       return end;
    }

private:
    bool satisfied(vector<int> &houses, vector<int> &heaters, int target) {
       
       int num_house = houses.size();
       int num_heater = heaters.size();
       for (int i = 0; i < num_house; ++i) {
           int closestDist = INT_MAX;

           //find the closest heaters
           int start = 0, end = num_heater - 1;
           while(start + 1 < end) {
               int mid = start + (end - start) / 2;
               if (heaters[mid] == houses[i]) {
                   closestDist = 0;
                   break;
               } else if (heaters[mid] < houses[i]) {
                   start = mid;
               } else {
                   end = mid;
               }
           }
           
           if (closestDist > 0) {
               closestDist = min(abs(houses[i] - heaters[start]), abs(houses[i] - heaters[end]));
               if (closestDist > target) {
                   return false;
               }
           }
       }
       return true;
    }
};c

解法2：

看了标准答案才发现，这道题其实就是找到所有单个house到离其最近的heater的最小值，然后返回所有这些最小值中的最大值即可。

class Solution {
public:
    /**
     * @param houses: positions of houses
     * @param heaters: positions of heaters
     * @return: the minimum radius standard of heaters
     */
    int findRadius(vector<int> &houses, vector<int> &heaters) {
       int num_house = houses.size();
       int num_heater = heaters.size();
       int radius = 0;
       sort(heaters.begin(), heaters.end());
       int minDist = 0;
   
       for (int i = 0; i < num_house; ++i) {
           int start = 0, end = num_heater - 1;
           int dist = INT_MAX;
           while(start + 1 < end) {
               int mid = start + (end - start) / 2;
               if (heaters[mid] == houses[i]) {
                   dist = 0;
                   break;
               } else if (heaters[mid] < houses[i]) {
                   start = mid;
               } else {
                   end = mid;
               }
           }
           
           if (dist > 0) {
               dist = min(abs(houses[i] - heaters[start]), abs(houses[i] - heaters[end]));
               minDist = max(minDist, dist);
           }
       }
       return minDist;
    }
};