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.翻译：
冬天来了！ 比赛期间你的第一份工作是设计一个具有固定半径的标准加热器来加热所有房屋。
现在，你被给定了水平上的房屋的位置和加热器的位置，找到最小的加热器半径以至于所有的房屋都能被这些加热器覆盖。
因此，你的输入将分别是房屋和加热器的位置，并且你的期待输出将是标准加热器的最小半径。
注意：1.你被给定的房屋和加热器的数量是正的并且不超过25000。
2.你被给定的房屋和加热器的位置是正的并且不超过10^9。
3.房屋是温暖的，只要房屋在加热器的加热半径范围内。4.所有的加热器都遵循你的半径标准，温暖的半径也是一样的。
例子1：
输入: [1,2,3],[2] 输出: 1 解释: 只有一个加热器在位置2，如果我们使用半径1为标准，那么所有的房屋都能被温暖。例子2:
输入: [1,2,3,4],[1,4] 输出: 1 解释: 两个加热器分别在位置1和4，如果我们使用半径1为标准，那么所有的房屋都能被温暖。思路：
先排序以确保房屋和加热器是有序的，然后给两个指针，按照顺序找到与每个屋子距离最近的加热器，记录其位置差，所有的位置差里面最大的那一个就是最小的加热器半径了。
C++ 代码（Visual studio 2017）：
#include "stdafx.h" #include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public: int findRadius(vector<int>& houses, vector<int>& heaters) { int result=0; sort(houses.begin(), houses.end()); sort(heaters.begin(), heaters.end()); int j = 0; for (int i = 0; i < houses.size(); i++) { while (j < heaters.size() - 1 && abs(heaters[j] - houses[i]) >= abs(heaters[j + 1] - houses[i])) { j++; } result = max(result, abs(heaters[j] - houses[i])); } return result; } }; int main() { Solution s; vector<int> houses = { 1,2,3,4 }; vector<int> heaters = { 1,4 }; int result; result = s.findRadius(houses, heaters); cout << result; return 0; }