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.
思路:首先如何使得每个station之间的最大距离最小,比如:两个station为[1, 9],间隔为8。要插入一个station使得最大距离最小,插入后应该为[1, 5, 9],最大间隔为4。如果插入后为[1, 6, 9], [1, 3, 9],它们的最大间隔分别为5, 6,不是最小。可以看出,对于插入k个station使得最大间隔最小的唯一办法是均分。
一种贪心的做法是,找到最大的gap,插入1个station,依此类推,但很遗憾,这种贪心策略是错误的。问题的难点在于我们无法确定到底哪两个station之间需要插入station,插入几个station也无法得知。用DP会内存超标MLE,用堆会时间超标TLE。
换个思路,如果假设知道了答案会怎么样?因为知道了最大间隔,所以如果目前的两个station之间的gap没有符合最大间隔的约束,就必须添加新的station来让它们符合最大间隔的约束,这样对于每个gap能够求得需要添加station的个数。如果需求数<=K,说明还可以进一步减小最大间隔,直到需求数>K。
解法1: 优先队列,每次找出gap最大的一个区间,然后新加入一个station,直到所有的k个station都被加入,此时最大的gap即为所求。采用优先队列,使得每次最大的gap总是出现在队首。空间复杂度是O(n),时间复杂度是O(klogn),其中k是要加入的新station的个数,n是原有的station个数。这种方法应该没毛病,但是TLE
解法:二分法,判定条件不是简单的大小关系,而是根据子函数。minmaxGap的最小值left = 0,最大值right = stations[n - 1] - stations[0]。每次取mid为left和right的均值,然后计算如果mimaxGap为mid,那么最少需要添加多少个新的stations,记为count。如果count > K,说明均值mid选取的过小,必须新加更多的stations才能满足要求,更新left的值;否则说明均值mid选取的过大,使得需要小于K个新的stations就可以达到要求,寻找更小的mid,使得count增加到K。如果假设stations[N- 1] - stations[0] = m,空间复杂度是O(1),时间复杂度是O(nlogm),可以发现与k无关。
Java:
public double minmaxGasDist(int[] stations, int K) { int n = stations.length; double[] gap = new double[n - 1]; for (int i = 0; i < n - 1; ++i) { gap[i] = stations[i + 1] - stations[i]; } double lf = 0; double rt = Integer.MAX_VALUE; double eps = 1e-7; while (Math.abs(rt - lf) > eps) { double mid = (lf + rt) /2; if (check(gap, mid, K)) { rt = mid; } else { lf = mid; } } return lf; } boolean check(double[] gap, double mid, int K) { int count = 0; for (int i = 0; i < gap.length; ++i) { count += (int)(gap[i] / mid); } return count <= K; }
Python:
class Solution(object): def minmaxGasDist(self, stations, K): """ :type stations: List[int] :type K: int :rtype: float """ def possible(stations, K, guess): return sum(int((stations[i+1]-stations[i]) / guess) for i in xrange(len(stations)-1)) <= K left, right = 0, 10**8 while right-left > 1e-6: mid = left + (right-left)/2.0 if possible(mid): right = mid else: left = mid return left
Python:
class Solution(object): def minmaxGasDist(self, st, K): """ :type stations: List[int] :type K: int :rtype: float """ lf = 1e-6 rt = st[-1] - st[0] eps = 1e-7 while rt - lf > eps: mid = (rt + lf) / 2 cnt = 0 for a, b in zip(st, st[1:]): cnt += (int)((b - a) / mid) if cnt <= K: rt = mid else: lf = mid return rt
Python:
class Solution(object): def minmaxGasDist(self, stations, K): """ :type stations: List[int] :type K: int :rtype: float """ stations.sort() step = 1e-9 left, right = 0, 1e9 while left <= right: mid = (left + right) / 2 if self.isValid(mid, stations, K): right = mid - step else: left = mid + step return mid def isValid(self, gap, stations, K): for x in range(len(stations) - 1): dist = stations[x + 1] - stations[x] K -= int(math.ceil(dist / gap)) - 1 return K >= 0
C++:
class Solution { public: double minmaxGasDist(vector<int>& stations, int K) { double left = 0, right = 1e8; while (right - left > 1e-6) { double mid = left + (right - left) / 2; if (helper(stations, K, mid)) right = mid; else left = mid; } return left; } bool helper(vector<int>& stations, int K, double mid) { int cnt = 0, n = stations.size(); for (int i = 0; i < n - 1; ++i) { cnt += (stations[i + 1] - stations[i]) / mid; } return cnt <= K; } };
C++:
class Solution { public: double minmaxGasDist(vector<int>& stations, int K) { double left = 0, right = 1e8; while (right - left > 1e-6) { double mid = left + (right - left) / 2; int cnt = 0, n = stations.size(); for (int i = 0; i < n - 1; ++i) { cnt += (stations[i + 1] - stations[i]) / mid; } if (cnt <= K) right = mid; else left = mid; } return left; } };
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