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LeetCode 2523. Closest Prime Numbers in Range(2025/3/7每日一题)

news 来源：原创 2025/6/10 16:58:55

标题：Closest Prime Number in Range

题目：

Given two positive integers left and right, find the two integers num1 and num2 such that:

left <= num1 < num2 <= right .
Both num1 and num2 are prime numbers.
num2 - num1 is the minimum amongst all other pairs satisfying the above conditions.

Return the positive integer array ans = [num1, num2]. If there are multiple pairs satisfying these conditions, return the one with the smallest num1 value. If no such numbers exist, return [-1, -1].

Example 1:

Input: left = 10, right = 19
Output: [11,13]
Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19.
The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19].
Since 11 is smaller than 17, we return the first pair.

Example 2:

Input: left = 4, right = 6
Output: [-1,-1]
Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.

Constraints:

1 <= left <= right <= 106

解题思路：本题的难点在如何能快速判断是否为质数。判断n是否为质数，则只需要判断其是否能整除2~sqrt(n)之间的任意一质数即可。为什么只需要判断到sqrt(n)呢？因为如果能整除，期中一个除数必然会落在2~sqrt(n)之间，如果都落在sqrt(n)之外，该数就大于n了。题中要求两个最近的质数，则输出肯定是相邻的两个质数。因此我们需要：

首先遍历left~right中的所有质数
遍历的同时记录相邻两个质数的距离
选取最近的两个质数输出

代码：

class Solution {
private: 
    vector<int> prim;
public:
    bool isPrim(int n) {
        for(int i = 0; i < prim.size() && prim[i] * prim[i] <= n; i++){
            if (n % prim[i] == 0) return false;
        }
        prim.push_back(n);
        return true;
    }
    vector<int> closestPrimes(int left, int right) {
        vector<int> res(2, -1);
        int last = 0, diff = INT_MAX;
        for(int i = 2; i <= right; i++){
            if (isPrim(i) && i >= left) {
                if (last != 0 && i - last < diff) {
                    diff = i - last;
                    res = {last, i};
                }
                last = i;
            }
        }
        return res;
    }
};

时间复杂度为O(N*log(N))，不是最优解

还可以用Sieve of Eratosthenes（埃拉托斯特尼筛法）来判断质数，时间复杂度为O(N*log(logN))