Leetcode 475. Heaters 加热器 解题报告

1 解题思想

这道题就是有一个加热器的地址列表，还有一个屋子列表，现在需要让你制定一个加热器的加热半径，使得所有的屋子都能被加热（加热器的位置就是加热器地址列表里的了）

做法也很简单，先排序，然后给两个指针，按照顺序找到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长的那一个就是最小的加热器半径了

2 原题

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: 1Explanation: 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: 1Explanation: 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.S

3 AC解

public class Solution {    public int findRadius(int[] houses, int[] heaters) {        Arrays.sort(houses);        Arrays.sort(heaters);        int n = houses.length;        int m = heaters.length;        int minimum = 0;        int j = 0;        for(int i = 0 ;i< n ;i++){            while( j < m - 1 && (Math.abs(heaters[j] - houses[i]) >= Math.abs(heaters[j+1] - houses[i])))                j ++;            //System.out.println(i+" "+j);            minimum = Math.max(minimum,Math.abs(heaters[j] - houses[i]));        }        return minimum;    }}