LeetCode 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. Note:

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.

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.

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给定房子和加热器的位置，问加热器的最小半径是所少才能使所有房子都在加热的范围之内。一个加热器的加热范围是一个圆，所有加热器的加热半径都是一样的。 要保证每个房子都在加热范围之内，且加热器的半径最小，则房子i肯定被紧邻i前后的两个加热器之一覆盖，其他加热器距离i的距离肯定比紧邻i前后的两个加热器远。比如x, i, y，i表示房子坐标，x和y分别表示紧邻i的两个加热器坐标，则覆盖i的最小半径应该是min(i-x,y-i)。 所以问题就转换为对每个房子坐标i，去加热器坐标数组中找到紧邻i前后的两个加热器坐标x和y，然后根据上述公式更新结果。为了便于查找，先对加热器坐标排序，然后直接二分查找就好了。 代码如下： [cpp] class Solution { public: int findRadius(vector<int>& houses, vector<int>& heaters) { sort(heaters.begin(), heaters.end()); int ans = 0; for (int i = 0; i < houses.size(); ++i) { int &pos = houses[i]; int r = INT_MAX; auto lb = lower_bound(heaters.begin(), heaters.end(), pos); if (lb > heaters.begin())r = min(r, pos – *(lb – 1)); auto ub = lower_bound(heaters.begin(), heaters.end(), pos); if (ub < heaters.end())r = min(r, *ub – pos); ans = max(ans, r); } return ans; } }; [/cpp] 本代码提交AC，用时89MS。 还可以自己写二分查找代码，如下： [cpp] class Solution { private: int my_lower_bound(vector<int>& v, int target) { int l = 0, r = v.size() – 1; while (l <= r) { int m = l + (r – l) / 2; if (target > v[m])l = m + 1; else r = m – 1; } return l; } public: int findRadius(vector<int>& houses, vector<int>& heaters) { sort(heaters.begin(), heaters.end()); int ans = 0, n = houses.size(), m = heaters.size(); for (int i = 0; i < n; ++i) { int &pos = houses[i]; int r = INT_MAX; int lb = my_lower_bound(heaters, pos); if (lb > 0)r = min(r, pos – heaters[lb – 1]); if (lb < m)r = min(r, heaters[lb] – pos); ans = max(ans, r); } return ans; } }; [/cpp] 本代码提交AC，用时76MS。]]>