为什么80%的码农都做不了架构师？>>> 
问题：
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: 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.
解决：
【注】参数中的两个数组分别表示房屋的位置和加热器的位置。现在需要让你指定一个加热器的加热半径，使得所有的屋子都能被加热。
①
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. 最小加热器半径radius = min(house - left, right - house)。所有house中距离的最大值就是minRadius.
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.MAX_VALUE。
public class Solution { // 31ms
public int findRadius(int[] houses, int[] heaters) {
Arrays.sort(heaters);
int minRadius = -1;
for (int house : houses) {
int index = Arrays.binarySearch(heaters, house);
if (index < 0) {
index = -(index + 1);
}
int radiusL = index - 1 >= 0 ? house - heaters[index - 1] : Integer.MAX_VALUE;//左侧半径
int radiusR = index < heaters.length ? heaters[index] - house : Integer.MAX_VALUE;//右侧半径
minRadius = Math.max(minRadius, Math.min(radiusL, radiusR));
}
return minRadius;
}
}
② 先排序，然后对heaters采用二分查找，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。
public class Solution { //25ms
public int findRadius(int[] houses, int[] heaters) {
Arrays.sort(houses);
Arrays.sort(heaters);
int left = 0;
int right = heaters.length - 1;
int minRadius = 0;
for (int i = 0;i < houses.length ;i ++ ) {
int cur = houses[i];
while(left < right && Math.abs(heaters[left + 1] - cur) <= Math.abs(heaters[left] - cur)){ //找到距离cur最近的加热器
left ++;
}
minRadius = Math.max(minRadius,Math.abs(heaters[left] - houses[i]));
}
return minRadius;
}
}
③ 进化版。
public class Solution { //18ms
public int findRadius(int[] houses, int[] heaters) {
Arrays.sort(houses);
Arrays.sort(heaters);
int left = 0;
int right = heaters.length;
int minRadius = 0;
for (int house : houses) {
int radius = 0;
while (left < right && house > heaters[left]) left ++;
if (left == 0 && house <= heaters[0]) radius = heaters[0]-house;
else if (left == right) radius = house - heaters[right - 1];
else {
radius = Math.min(heaters[left] - house, house - heaters[left - 1]);
}
if (radius > minRadius) minRadius = radius;
}
return minRadius;
}
}