Finding palindrome numbers

#include <stdio.h> #include <math.h> int is_Palindrome(int num1, int num2) { if (num1 > num2) { //printf("base condition\n"); return 0; } else { if (num1 == reverse(num1)) printf("%d ", num1); num1 = num1 + 1; } is_Palindrome(num1, num2); } int reverse(int num1) { int dig = (int)log10(num1); if (num1 == 0) { // base condition return 0; } else return ((num1 % 10 * pow(10, dig)) + reverse(num1 / 10)); } int main(int argc, char const *argv[]) { int num1, num2; printf("enter two numbers:\t"); scanf("%d%d", &num1, &num2); is_Palindrome(num1, num2); return 0; }

//Finding palindrome numbers

// first convert to string (slow)
fn is_palindrome1(num: i32) -> bool {
    return num.to_string().chars().rev().collect::<String>() == num.to_string();
}

// variation using zip
fn is_palindrome2(num: i32) -> bool {
    let s = num.to_string();
    for (i1, i2) in s.chars().zip(s.chars().rev()) {
        if i1 != i2 {
            return false;
        }
    }
    return true;
}

// divide by 10 (fast)
fn is_palindrome3(num: i32) -> bool {
    let mut reversed = 0;
    let mut number = num;
    while number > 0 {
        reversed = reversed * 10 + number % 10;
        number = number / 10;
    }
    return reversed == num || num == number / 10;
}

fn main() {
    for i in 700..750 {
        if is_palindrome1(i) {
            println!("Pal1: {}", i);
        }
        if is_palindrome2(i) {
            println!("Pal2: {}", i);
        }        
        if is_palindrome3(i) {
            println!("Pal3: {}", i);
        }
    }
}

Posted by: Guest on September-06-2021

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