786. K-th Smallest Prime Fraction

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.
• Each A[i] will be between 1 and 30000.
• K will be between 1 and A.length * (A.length - 1) / 2.

Approach #1: brute force.[Time Limit Exceeded]

class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& A, int K) {
        int n = A.size();
        vector<pair<int, int>> temp;
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            for (int j = i+1; j < n; ++j) {
                temp.push_back({A[i], A[j]});
            }
        }
        sort(temp.begin(), temp.end(), cmp);
        ans.push_back(temp[K-1].first);
        ans.push_back(temp[K-1].second);
        return ans;
    }
    
    static bool cmp (pair<int, int> a, pair<int, int> b) {
        return (double)a.first/a.second < (double)b.first/b.second;
    }
};

 

Approach #2: Binary Search.

class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& A, int K) {
        int n = A.size();
        double l = 0.0, r = 1.0;
        while (l < r) {
            double m = (l + r) / 2;
            double max_f = 0.0;
            int total = 0;
            int p, q;
            int j = 1;
            for (int i = 0; i < n-1; ++i) {
                while (j < n && A[i] > m * A[j]) ++j;
                total += (n - j);
                if (j == n) break;
                double f = static_cast<double>(A[i]) / A[j];
                if (f > max_f) {
                    p = i;
                    q = j;
                    max_f = f;
                }
            }
            if (total == K) return {A[p], A[q]};
            else if (total < K) l = m;
            else r = m;
        }
        return {};
    }
};

Runtime: 12 ms, faster than 83.88% of C++ online submissions for K-th Smallest Prime Fraction.

 

Analysis:

1. why use static_cast<>?

In short:

1. static_cast<>() gives you a compile time checking ability, C-Style cast doesn't.
2. static_cast<>() can be spotted easily anywhere inside a C++ source code; in contrast, C_Style cast is harder to spot.
3. Intentions are conveyed much better using C++ casts.

More Explanation:

The static cast performs conversions between compatible types. It is similar to the C-style cast, but is more restrictive. For example, the C-style cast would allow an integer pointer to point to a char.

char c = 10;       // 1 byte
int *p = (int*)&c; // 4 bytes

Since this results in a 4-byte pointer pointing to 1 byte of allocated memory, writing to this pointer will either cause a run-time error or will overwrite some adjacent memory.

*p = 5; // run-time error: stack corruption

In contrast to the C-style cast, the static cast will allow the compiler to check that the pointer and pointee data types are compatible, which allows the programmer to catch this incorrect pointer assignment during compilation.

int *q = static_cast<int*>(&c); // compile-time error

 

2. In this way we make m as a flag (range is from 0 to 1),  we can count the numbers which elements less than m. when if m == K we return {A[p], A[q]} , else if m > K => r = m; else l = m

we can use a matrix to instore the A[i] / A[j], but in order to reduce the time complexity  we can use under code to realize.

　　
for (int i = 0; i < n-1; ++i)
　　while (j < n && A[i] > m * A[j]) ++j;

for every loop j can be using repeatedly, because when i increase, in order to make A[i] > m * A[j], j must bigger than last loop.

 

转载于:https://www.cnblogs.com/ruruozhenhao/p/9936650.html