leetcode 906. Super Palindromes

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

Now, given two positive integers L and R (represented as strings), return the number of superpalindromes in the inclusive range [L, R].

 

Example 1:

Input: L = "4", R = "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.

 

Note:

1. 1 <= len(L) <= 18
2. 1 <= len(R) <= 18
3. L and R are strings representing integers in the range [1, 10^18).
4. int(L) <= int(R)

解题思路（思路进阶过程）；

1. 遍历数字1到sqrt(stoll(R)) , 判断自身是否回文数且平方数是否回文数；

TLE ， 因为有很多数自身不是回文数，产生无效判断 ；

class Solution {
public:
    int superpalindromesInRange(string L, string R) 
    {
        int res = 0 ; 
        long long lnum = stoll(L) , rnum = stoll(R) ;
        for(long long i = 1 ; i <= (long long)sqrt(rnum) ; ++i)
        {
            if(!ispalin(i)) continue ;
            long long square = i * i ;
            if(ispalin(square) && square >= lnum && square <= rnum) res++ ;
        }
        
        return res ;
    }
    
    bool ispalin(long long i)
    {
        string s = to_string(i) ;
        if(s.size() == 1) return true ;
        for(int i = 0 , j = s.size() - 1 ; i <= j ; ++i , --j)
        {
            if(s[i] != s[j]) return false ;
        }
        return true ;
    }
};

2 . 构造sz位数的回文数 ， 范围从[ to_string(sqrt(left)).size() , min(9 , to_string(sqrt(right)).size())] , 因为本身是回文数，如果其平方数也是回文数，则res++ ；

class Solution {
public:
    int superpalindromesInRange(string L, string R) 
    {

        lnum = stoll(L) ;
        rnum = stoll(R) ;
        int res = 0 ; 
        long long sqrtr = sqrt(rnum) + 1 , sqrtl = sqrt(lnum) ;
        string sqrtrs = to_string(sqrtr) , sqrtls = to_string(sqrtl) ;
        for(int sz = sqrtls.size() ; sz <= min(9 ,(int)sqrtrs.size()) ; ++sz)
        {
            string s(sz , '.') ;
            DFS(0 , sz , s , res) ;
        }
        return res ;
    }
    
    bool ispalin(string s)
    {
        if(s.empty() || s.size() == 1) return true ;
        for(int i = 0 , j = s.size() - 1 ; i <= j ; ++i , --j)
        {
            if(s[i] != s[j]) return false  ;
        }
        return true ;
    }
    
    void DFS(int pos , int sz , string& s , int& res)
    {
        if(pos > sz - 1 - pos)
        {
            long long num = stoll(s) ;          
            long long squanum = num * num ;
            if( squanum < lnum || squanum > rnum) return ;
            string squas = to_string(squanum) ;
            if(ispalin(squas))
            {
                res++ ;
            }
            return ;
        }
        
        for(int i = (pos == 0 ? 1 : 0) ; i < 10 ; ++i)
        {
            s[pos] = '0' + i ;
            s[sz - 1 - pos] = '0' + i ;
            DFS(pos + 1 , sz , s , res) ;
        }
    }
    
private :
    long long lnum , rnum ;
};