Heaters

https://leetcode.com/problems/heaters/description/

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 separately, 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.

Thoughts

刚开始以为给的是排好序了的，实际上并没有。

原来想单独分析houses之中和heaters之中的距离，然后再比较二者之间的距离，后失败。

实际上对每个house, 与它最近heater之间的距离就是它所需的最短辐射长度。我们要找针对所有houses的最短辐射长度，就取每个house的所需最短辐射长度最大值即可。

那么对给定的house如何找最近的heater? 把heaters排个序然后二分法即可。

Code

class Solution {
    public int findRadius(int[] houses, int[] heaters) {
        Arrays.sort(heaters);
        int max = 0;
        for (int house : houses) {
            int start = 0, end = heaters.length;
            while (start < end) {
                int mid = start + (end - start) / 2;
                if (heaters[mid] < house) {
                    start = mid + 1;
                } else {
                    end = mid;
                }
            }
            int dist = start < heaters.length ? Math.abs(heaters[start] - house) : Integer.MAX_VALUE;
            dist = start > 0 ? Math.min(dist, Math.abs(heaters[start - 1] - house)) : dist;
            max = Math.max(max, dist);
        }

        return max;
    }
}

Analysis

排序的时间复杂读是O(nlgn), 再加上找raius的O(mlgn).