475. Heaters - Easy

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.

 

ref:  http://www.cnblogs.com/grandyang/p/6181626.html

用binary search找到第一个 >= 当前house位置的数，如果heaters中存在这个数，可以求其与house位置的差值，如果这个数不是heater的第一个元素，可以求出house与这个数前一个数的差值并取其中较小的为能cover当前house的最小半径

time: O(), space: O(1)

class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        int res = 0, n = heaters.length;
        Arrays.sort(heaters);
        for(int house : houses) {
            int left = 0, right = n;
            while(left < right) {
                int mid = left + (right - left) / 2;
                if(heaters[mid] < house) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            int d1 = right == n ? Integer.MAX_VALUE : heaters[right] - house;
            int d2 = right == 0 ? Integer.MAX_VALUE : house - heaters[right - 1];
            res = Math.max(res, Math.min(d1, d2));
        }
        return res;
    }
}