0906. Super Palindromes (H)

Super Palindromes (H)

题目

Let's say a positive integer is a super-palindrome if it is a palindrome, and it is also the square of a palindrome.

Given two positive integers left and right represented as strings, return the number of super-palindromes integers in the inclusive range [left, right].

Example 1:

Input: left = "4", right = "1000"
Output: 4
Explanation: 4, 9, 121, and 484 are superpalindromes.
Note that 676 is not a superpalindrome: 26 * 26 = 676, but 26 is not a palindrome.

Example 2:

Input: left = "1", right = "2"
Output: 1

Constraints:

• 1 <= left.length, right.length <= 18
• left and right consist of only digits.
• left and right cannot have leading zeros.
• left and right represent integers in the range [1, 10^18].
• left is less than or equal to right.

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题意

定义一个超级回文数n，具有如下性质：n本身是回文数，n的平方根同样是回文数。找到指定范围内所有的超级回文数。

题意

直接构造所有的1e10之内的回文数，判断它们的平方是否在范围内且同样为回文数。注意回文数的长度既可能是奇数也可能是偶数。

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思路

Java

class Solution {
    private int count = 0;

    public int superpalindromesInRange(String left, String right) {
        long a = Long.parseLong(left), b = Long.parseLong(right);
        count = 0;

        dfs("", a, b);
        for (int i = 0; i <= 9; i++) {
            dfs(String.valueOf(i), a, b);
        }

        return count;
    }

    private void dfs(String s, long left, long right) {
        if (s.length() > 9) return;


        if (s.length() > 0) {
            long x = Long.parseLong(s);
            if (x * x > right) return;
            if (s.charAt(0) != '0' && x * x >= left && isPalindrome(x) && isPalindrome(x * x)) count++;
        }

        for (int i = 0; i <= 9; i++) {
            dfs(i + s + i, left, right);
        }
    }

    private boolean isPalindrome(long x) {
        char[] s = String.valueOf(x).toCharArray();
        int i = 0, j = s.length - 1;
        while (i < j) {
            if (s[i++] != s[j--]) return false;
        }
        return true;
    }

    public static void main(String[] args) {
        Solution s = new Solution();
        s.superpalindromesInRange("4", "1000");
    }
}