一、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, 1018 - 1].
left is less than or equal to right.

二、题解

class Solution {
public:bool isPalindrome(long long num){long offset = 1;while(num / offset >= 10){offset *= 10;}while(num){if(num / offset != num % 10) return false;num = (num % offset) / 10;offset /= 100;}return true;}long long oddEnlarge(long long seed){long long res = seed;seed /= 10;while(seed){res = res * 10 + seed % 10;seed /= 10;}return res;}long long evenEnlarge(long long seed){long long res = seed;while(seed){res = res * 10 + seed % 10;seed /= 10;}return res;}bool check(long long num,long l,long r){return num >= l && num <= r && isPalindrome(num);}int superpalindromesInRange(string left, string right) {long long l = stol(left);long long r = stol(right);long long limit = (long long)sqrt((double)r);long long seed = 1;long long num = 0;int res = 0;do{//偶数长度回文数num = evenEnlarge(seed);if(num <= (long long)sqrt(double(LONG_LONG_MAX)) && check(num * num,l,r)) res++;//奇数长度回文数num = oddEnlarge(seed);if(num <= (long long)sqrt(double(LONG_LONG_MAX)) && check(num * num,l,r)) res++;seed++;}while(num < limit);return res;}
};