479. Product of Maximum Palindromic Numbers: Enumeration Problems

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Topic description

This is 479. Maximum Palindromic Product on LeetCode on Hard difficulty .

Tag : "enumeration", "math"

given an integer $n$ , the return can be expressed as two $n$  -bit integers.

Because the answer can be very large, return it for $1337$ Extra.

Example 1:

输入：n = 2

输出：987

解释：99 x 91 = 9009, 9009 % 1337 = 987
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Example 2:

输入： n = 1

输出： 9
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hint:

• $1 <= n <= 8$

Enumeration + Math

for digits $n$ , the number of digits in the product is either $2 * n$ , either $2 * n - 1$。

when digital $n > 1$ , we can always $2 * n$ .

Using the characteristics of the palindrome, we only need to enumerate the first half of the palindrome (the second half is uniquely determined), and we only need to follow the "large to small" when enumerating the first half to ensure that the found The first legal value is the largest number, and for a digit is $n$The maximum number of$n\; is$ $10^n - 1$。

具体的，当枚举到回文串的前半部分 $i$ 时，我们利用回文串特性构造出具实际的回文数值 $nums$，随后检查 $nums$ 能否分解成数位为 $n$ 的数对 $(a, b)$，利用乘法具有交换律，我们只需要枚举数对中的较大数即可。

代码：

class Solution {
    public int largestPalindrome(int n) {
        if (n == 1) return 9;
        int max = (int) Math.pow(10, n) - 1;
        for (int i = max; i >= 0; i--) {
            long num = i, t = i;
            while (t != 0) {
                num = num * 10 + (t % 10);
                t /= 10;
            }
            for (long j = max; j * j >= num; j--) {
                if (num % j == 0) return (int)(num % 1337);
            }
        }
        return -1;
    }
}
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• 时间复杂度：枚举回文串的前半部分复杂度为 $O(10^n)$；检查回文串能否被分解复杂度为 $O(10^n)$。整体复杂度为 $O(10^{2n})$
• 空间复杂度： $O(1)$

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