Minimize Max Distance to Gas Station

On a horizontal number line, we have gas stations at positions stations[0], stations[1], ..., stations[N-1], where N = stations.length.

Now, we add K more gas stations so that D, the maximum distance between adjacent gas stations, is minimized.

Return the smallest possible value of D.

Example:

Input: stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 9
Output: 0.500000

Note:

1. stations.length will be an integer in range [10, 2000].
2. stations[i] will be an integer in range [0, 10^8].
3. K will be an integer in range [1, 10^6].
4. Answers within 10^-6 of the true value will be accepted as correct.

思路：就是如果加入0個station，那麼上限就是現在的max adjcent distance， 如果加入無窮多個station，那麼下限就是distance = 0;

那麼就是要找到恰當的dis 使得加入的station是K，那麼就是個二分答案的binary search，輔助方程是求給你distance，求要加入多少個stations. 求station 個數就用station[i + 1] - station[i] / dis

注意double的判斷，不用用start + 1 < end，要用eps = 1e-5 , start + eps < end來判斷；

class Solution {
    public double minmaxGasDist(int[] stations, int K) {
        if(stations == null || stations.length == 0) {
            return 0.0;
        }
        
        double maxdis = 0;
        for(int i = 0; i < stations.length - 1; i++) {
            maxdis = Math.max(maxdis, stations[i + 1] - stations[i]);
        }
        
        double start = 0;  double end = maxdis;
        double eps = 1e-5;
        while(start + eps < end) {
            double mid = start + (end - start) / 2;
            if(needstation(stations, mid) > K) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if(needstation(stations, start) > K) {
            return end;
        }
        return start;
    }
    
     private int needstation(int[] stations, double dis) {
        int count = 0;
        int i = 0;
        while(i < stations.length - 1) {
            count += (stations[i + 1] - stations[i]) / dis;
            i++;
        }
        return count;
    }
}