On a horizontal number line, we have gas stations at positions stations[0], stations[1], ..., stations[N-1]
, where N = stations.length
.spa
Now, we add K
more gas stations so that D, the maximum distance between adjacent gas stations, is minimized.code
Return the smallest possible value of D.blog
Example:it
Input: stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 9 Output: 0.500000
Note:io
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.
貪心確定不對,參考discuss,假設一個結果,而後對結果進行二分逼近。class
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; //重點理解代碼,d_i / (cnt_i + 1) <= mid } if (cnt > K) { left = mid; } else { right = mid; } } return mid; }