[LeetCode] 774. Minimize Max Distance to Gas Station 最小化加油站间的最大距离
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.lengthwill be an integer in range[10, 2000].stations[i]will be an integer in range[0, 10^8].Kwill be an integer in range[1, 10^6].- Answers within
10^-6of 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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