LeetCode每日一题(1362. Closest Divisors)

Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.

Return the two integers in any order.

Example 1:

Input: num = 8
Output: [3,3]

Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.

Example 2:

Input: num = 123
Output: [5,25]

Example 3:

Input: num = 999
Output: [40,25]

Constraints:

• 1 <= num <= 10^9

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最优解当然是 n² = num + 1 或者 n² = num + 1, 因为 1 <= num <= 1000000000, 那我们的 n 自然不会大于 40000, 有了这个条件托底， 我们就算用最简单的方法从 40000 一路试到 1 也可以, 当然， 最好是把 n 先判断出来

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impl Solution {
    
    pub fn closest_divisors(num: i32) -> Vec<i32> {
    
        if num == 1 {
    
            return vec![1, 2];
        }
        let num = num as i64;
	// 指数
        let mut e = 1u32;
        loop {
    
            if 2i64.pow(e) >= num {
    
                break;
            }
            e += 1;
        }
	// 计算出n的上限
        let n = 2i64.pow(e / 2 + e % 2);
        let mut a = i64::MAX;
        let mut b = 0;
        for v in (1..=n).rev() {
    
            if (num + 1) % v == 0 {
    
                if ((num + 1) / v - v).abs() < (a - b).abs() {
    
                    a = (num + 1) / v;
                    b = v;
                }
            }
            if (num + 2) % v == 0 {
    
                if ((num + 2) / v - v).abs() < (a - b).abs() {
    
                    a = (num + 2) / v;
                    b = v;
                }
            }
        }
        vec![a as i32, b as i32]
    }
}