LeetCode | 475. Heaters 循环技巧题

Winteris coming! Your first job during the contest is to design a standard heaterwith fixed warm radius to warm all the houses.

Now,you are given positions of houses and heaters on a horizontal line, find outminimum 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 yourexpected output will be the minimum radius standard of heaters.

Note:

1.      Numbers of houses and heaters you aregiven are non-negative and will not exceed 25000.

2.      Positions of houses and heaters youare 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 radiusstandard and the warm radius will the same.

Example1:

Input: [1,2,3],[2]

Output: 1

Explanation: The only heater was placed in theposition 2, and if we use the radius 1 standard, then all the houses can bewarmed.

Example2:

Input: [1,2,3,4],[1,4]

Output: 1

Explanation: The two heater was placed in theposition 1 and 4. We need to use radius 1 standard, then all the houses can bewarmed.

简单题，将两个数组排序，之后不断的求house数组里面的每一个房子离他最近的heater的距离，之后取所有这样的距离里面最大的一个就是我们要求的最短的散热范围

class Solution {

public:

intfindRadius(vector<int>& houses, vector<int>& heaters) {

      sort(houses.begin(), houses.end());

      sort(heaters.begin(), heaters.end());

 

      int index = 0; int m = -1;

      for (int i = 0; i < houses.size(); i++)

      {

            while (index <heaters.size()&& heaters[index] < houses[i]) index++;

            if (index == heaters.size()) m =max(m, houses[i] - heaters[index - 1]);

            else if(index==0)

            {

                  m = max(m, heaters[0] -houses[i]);

            }

            else

            {

                  m = max(m,min(houses[i]-heaters[index-1],heaters[index]-houses[i]));

            }

      }

      return m;

}

};