LeetCode Heaters

原题链接在这里：https://leetcode.com/problems/heaters/#/description

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:

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.

题解：

对于每一个house在sorted heaters中用Arrays.binarySearch找.

若是没有找到会return应该插入位置的负数. e.g. 在[1,2,3]中找5, 会return -4. 所以要算出实际插入的index, index = -(-4+1) = 3.

然后house - heaters[index-1]计算最近左侧heater的距离. heaters[index] - house计算最近右侧heater距离, 如果house本身就有heater时, heaters[index]就是house, 最近右侧距离等于0. 取左右中较小值为离为最近的heater距离.

所有house中距离的最大值就是res.

Note: 若index落在了heaters的首位, 左侧距离就是Integer.MAX_VALUE. 落在末位, 右侧距离就是Integer.MAC_VALUE. 

Time Complexity: O(nlogm). n = houses.length, m = heaters.length, sort用了mlogm, 但m肯定比n小.

Space: O(1).

AC Java:

public class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        int res = Integer.MIN_VALUE;
        Arrays.sort(heaters);
        for(int house : houses){
            int index = Arrays.binarySearch(heaters, house);
            if(index < 0){
                index = -(index+1);
            }
            
            int distLeft = index>=1 ? house-heaters[index-1] : Integer.MAX_VALUE;
            int distRight = index<heaters.length ? heaters[index]-house : Integer.MAX_VALUE;
            res = Math.max(res, Math.min(distLeft, distRight));
        }
        return res;
    }
}