A sorted list A
contains 1, plus some number of primes. Then, for every p < q in the list, we consider the fraction p/q.
What is the K
-th smallest fraction considered? Return your answer as an array of ints, where answer[0] = p
and answer[1] = q
.
Examples: Input: A = [1, 2, 3, 5], K = 3 Output: [2, 5] Explanation: The fractions to be considered in sorted order are: 1/5, 1/3, 2/5, 1/2, 3/5, 2/3. The third fraction is 2/5. Input: A = [1, 7], K = 1 Output: [1, 7]
Note:
A
will have length between 2
and 2000
.A[i]
will be between 1
and 30000
.K
will be between 1
and A.length * (A.length + 1) / 2
.我想着用 priority queue做,但是怎么做都TLE了。看看大神的解法吧。
This solution probably doesn’t have the best runtime but it’s really simple and easy to understand.
Says if the list is [1, 7, 23, 29, 47]
, we can easily have this table of relationships
1/47 < 1/29 < 1/23 < 1/7
7/47 < 7/29 < 7/23
23/47 < 23/29
29/47
So now the problem becomes “find the kth smallest element of (n-1) sorted list”
Following is my implementation using PriorityQueue, running time is O(max(n,k) * logn), space is O(n):
public int[] kthSmallestPrimeFraction(int[] a, int k) {
int n = a.length;
// 0: numerator idx, 1: denominator idx
PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
int s1 = a[o1[0]] * a[o2[1]];
int s2 = a[o2[0]] * a[o1[1]];
return s1 - s2;
}
});
for (int i = 0; i < n-1; i++) {
pq.add(new int[]{i, n-1});
}
for (int i = 0; i < k-1; i++) {
int[] pop = pq.remove();
int ni = pop[0];
int di = pop[1];
if (pop[1] - 1 > pop[0]) {
pop[1]--;
pq.add(pop);
}
}
int[] peek = pq.peek();
return new int[]{a[peek[0]], a[peek[1]]};
}
关注后续 discuss 区的优秀解答。 https://leetcode.com/problems/k-th-smallest-prime-fraction/discuss/