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:
-
stations.length
will be an integer in range[10, 2000]
. -
stations[i]
will be an integer in range[0, 10^8]
. -
K
will be an integer in range[1, 10^6]
. - Answers within
10^-6
of the true value will be accepted as correct.
贪心肯定不对,参考discuss,假设一个结果,然后对结果进行二分逼近。
public double minmaxGasDist(int[] stations, int K) {
int LEN = stations.length;
double left = 0, right = stations[LEN - 1] - stations[0], mid = 0; while (right >= left + 0.000001) {
mid = right - (right - left) / 2;
int cnt = 0;
for (int i = 0; i < LEN - 1; i++) {
cnt += Math.ceil((stations[i + 1] - stations[i]) / mid) - 1; //重点理解代码,d_i / (cnt_i + 1) <= mid
}
if (cnt > K) {
left = mid;
} else {
right = mid;
}
}
return mid;
}