Palindrome Number - LC9

Palindrome Number - LC9

Given an integer x, return true if x is a palindrome, and false otherwise.

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Table of contents

Question

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Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Answer

class Solution {
public:
    bool isPalindrome(int x) {
        int oldNum=x;
        int newNum=0;
        while(0<x){
            int temp=x%10;
            newNum=newNum*10+temp;
            x=x/10;
        }
        if (oldNum==newNum){
            return true;
        }
        else{
            return false;
        } 
    }
};

Here we will reverse the number and check that the given number is equal to the new number. First, we need to store the old number in a variable as the variable x will be 0 at last, so we are using oldNum it as a variable to remember our given number. The variable newNum is where our reverse number will be stored.

Now we will use a while loop and check if the x is greater than 0 and run it till the x become 0. The variable temp store the last number of the x, eg., given the number 121 then 121%10=1, 1 is stored in the temp . After we get the temp then we move to the next line, here we are multiplying the newNum with 10 and adding the last number which we stored in temp eg., newNum = 0*10+1 -> newNum = 1 and then we will remove the last number from the x by dividing x by 10. This loop continues till the x become 0.

Till this point new will get our reverse number which is stored in, now we just need to compare this with the oldNum, if the number is the same then return true or else false.

Input : 121

x = 121
oldNum = 121
newNum = 0
Check if x is greater than 0 (0<121) โœ…
    temp = x % 10 = 1 
    newNum = (newNum * 10) + temp = (0*10) + 1 = 0 + 1 = 1
    x = x / 10 = 121 / 10 = 12

x = 12
oldNum = 121
newNum = 1
Check if x is greater than 0 (0<12) โœ…
    temp = x % 10 = 12 % 10 = 2
    newNum = (newNum * 10) + temp = (1*10) + 2 = 10 + 2 = 12
     x = x / 10 = 12 / 10 = 1

x = 1
oldNum = 121
newNum = 12
Check if x is greater than 0 (0<1) โœ…
    temp = x % 10 = 1 % 10 = 1
    newNum = (newNum * 10) + temp = (12*10) + 1 = 120 + 1 = 121
    x = x / 10 = 1 / 10 = 0

x = 0
oldNum = 121
newNum = 121
Check if x is greater than 0 (0<0) โŒ

check if oldNum is equal to newNum (121 == 121) โœ…
    return true

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