LWC 72: 786. K-th Smallest Prime Fraction

LWC 72: 786. K-th Smallest Prime Fraction

传送门：786. K-th Smallest Prime Fraction

Problem:

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 answer1 = 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.

Each A[i] will be between 1 and 30000.

K will be between 1 and A.length * (A.length + 1) / 2.

思路1：

一种聪明的做法，如果A = [1, 7, 23, 29, 47]，那么有：

1/47   < 1/29    < 1/23 < 1/7
7/47   < 7/29    < 7/23
23/47  < 23/29
29/47

可以采用优先队列的方法，把第一列先送入队列，因为他们一定是每行的最小，选出最小的之后，加入该行的后续元素。直到找到K-1个最小的元素。

代码如下：

public int[] kthSmallestPrimeFraction(int[] a, int k) {
int n = a.length;
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.poll();
if (pop[1] - 1 > pop[0]) {
pop[1]--;
pq.offer(pop);
}
}

int[] peek = pq.peek();
return new int[]{a[peek[0]], a[peek[1]]};
}

思路2：

二分，先找到这个第k小的数，接着再小范围暴力搜索，居然能过。

代码如下：

public int[] kthSmallestPrimeFraction(int[] a, int K) {
double low = 0, high = 1;
double eps = 1e-8;
int n = a.length;
for(int rep = 0; rep < 50; ++rep){
double x = low + (high-low) / 2;
int num = 0;
for(int i = 0;i < n;i++){
int ind = Arrays.binarySearch(a, (int)(x * a[i]));
if(ind < 0) ind = -ind - 2;
num += ind + 1;
}
if(num >= K){
high = x;
}else{
low = x;
}
}

for (int i = 0; i < n; ++i) {
double find = a[i] * high;
int idx = Arrays.binarySearch(a, (int)find);
if (idx < 0) idx = -idx;
for (int j = -1; j <= 1; ++j) {
if (j + idx >= 0 && j + idx < n && Math.abs(a[j + idx] - find) < eps) {
return new int[] {a[j + idx], a[i]};
}
}
}
return new int[]{-1, -1};
}