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.
也是用了二分查找,找的是中位数。
Runtime:28ms Beats: 81.77%
class Solution {
public:
double minmaxGasDist(vector<int>& stations, int K) {
double left = 0.0, right = 1000000.0, mid = 0.0;
while (left + 0.0000001 < right) {
mid = left + (right - left) / 2.0;
int cnt = 0;
for (int i = 0; i < stations.size() - 1; i++) {
cnt += (stations[i + 1] - stations[i]) / mid;
}
if (cnt <= K) right = mid;
else left = mid;
}
return left;
}
};