Leetcode 475(Java)

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.

题目如上，不作翻译了~最开始只看了第一个例题以后自信满满开始思考写出代码以后run code通过，submit却不行。

public class Solution {    public int findRadius(int[] houses, int[] heaters) {        int max = 0;        int com = 0;        for(int i=0;i<heaters.length;i++){            int x = heaters[i];            for(int j=0;j<houses.length;j++){                int y = houses[j];                com = Math.abs(x-y);                if(com>max)max = com;            }        }        return max;    }}

是的。。。我把Heaters看成了单个元素，发现错误后重新改正了结题的思路。这道题目要用到排序，然后用两个指针来确定每个屋子到其最近的加热器的距离，记录下来以后取最大值作为加热器的半径。

例如房子坐标输入{1,2,3,4,5,6}
热水器坐标输入{2,4}
首先拿出房子的第0号单元，即数值1，先与热水器的第0号单元数值2比较，发现绝对差值为1，再与热水器的第1号单元4比较发现绝对差值为3，并不比1小，因此对于房子的第0号单元而言，离它最近的热水器坐标所需半径为1。依次类推，具体代码实现如下：
public class Solution {
public int findRadius(int[] houses, int[] heaters) {
Arrays.sort(houses);
Arrays.sort(heaters);
int minrad = 0;
int j =0;
for(int i=0;i<houses.length;i++){
while(j<(heaters.length-1) && Math.abs(houses[i]-heaters[j])>=Math.abs(houses[i]-heaters[j+1])){
j++;
}
minrad = Math.max(minrad,Math.abs(houses[i]-heaters[j]));
}
return minrad;
}
}