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.
冬天来了！您在比赛期间的第一份工作是设计一个固定温暖半径的标准加热器，以加热所有房屋。
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.
现在，您可以在水平线上获得房屋和加热器的位置，找出加热器的最小半径，以便所有房屋都能被这些加热器覆盖。
So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
因此，您的输入将分别是房屋和加热器的位置，您的预期输出将是加热器的最小半径标准。
Note:
- Numbers of houses and heaters you are given are non-negative and will not exceed 25000.您给出的房屋和加热器的数量是非负的，不会超过25000。
- Positions of houses and heaters you are given are non-negative and will not exceed 10^9.您给出的房屋和加热器的位置是非负的，不会超过10 ^ 9。
- As long as a house is in the heaters‘ warm radius range, it can be warmed.只要房屋处于加热器的温暖半径范围内，就可以加热。
- All the heaters follow your radius standard and the warm radius will the same.所有加热器都遵循半径标准，温暖半径也是如此。
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.唯一的加热器放置在位置2，如果我们使用半径1标准，那么所有的房屋都可以加热。
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.两个加热器放置在位置1和4.我们需要使用半径1标准，然后所有房屋都可以加热。
解答：
方法一：
1 class Solution {
2 public int findRadius(int[] houses, int[] heaters) {
3 Arrays.sort(heaters);
4 int res=0,n=heaters.length;
5
6 for(int house:houses){
7 int left=0,right=n;
8 while(left<right){
9 int mid=left+(right-left)/2;
10 if(house>heaters[mid])
11 left=mid+1;
12 else
13 right=mid;
14 }
15
16 int distance1=(right==n) ? Integer.MAX_VALUE:heaters[right]-house;
17 int distance2=(right==0) ? Integer.MAX_VALUE:house-heaters[right-1];
18 res=Math.max(res,Math.min(distance1,distance2));
19 }
20
21 return res;
22 }
23 }
详解：
先对heaters排序，然后用二分搜索法迅速找到heaters中第一个大于等于当前house位置的数，如果这个数存在，可以算出它和house的差值
如房屋house处于位置5，离它最近的heater为3和9，那么5-3=2,9-5=4，radis=2即可满足要求。
原文地址：https://www.cnblogs.com/chanaichao/p/9662869.html