Leetcode-9-palindromes (simple)

9. Palindrome number （ Simple ）

Give you an integer x , If x Is a palindrome integer , return true ; otherwise , return false .

Palindrome number refers to positive order （ From left to right ） Reverse order （ From right to left ） Read all the same integers . for example ,121 It's palindrome. , and 123 No .

** Advanced ：** Can you solve this problem without converting integers into strings ？

Example 1：

 Input ：x = 121
 Output ：true

Example 2：

 Input ：x = -121
 Output ：false
 explain ： Read left to right ,  by  -121 .  Read right to left ,  by  121- . So it's not a palindrome number .

Ideas ：

1. Negative numbers are not palindromes
2. Determine whether a positive number is a palindrome number , Judge whether the numbers before and after reversal are equal
   1. Convert a number into a string and judge
   2. Do not convert string , utilize %10 and /10, Reverse the number , Then judge whether it is equal or not

（1）C++

A time-consuming version of a string

class Solution {
public:
    bool isPalindrome(int x) {
        if(x<0)
            return false;
        string s = to_string(x);
        string s1 = s;
        reverse(s1.begin(),s1.end());
        if(s1==s)
            return true;
        else
            return false;
    }
};

Take advantage of a short, time-consuming version of string pointers

class Solution {
public:
    bool isPalindrome(int x) {
        if(x<0)
            return false;
        string s = to_string(x);
        bool temp=true;
        int i =0 , j=s.size()-1;
        while(i<j){
            if(s[i]!=s[j]){
                temp = false;
                break;
            }
            else{
                i++;
                j--;
            }
        }
        return temp;
    }
};

(2)C++（ No strings 、 Compare the first and last digits of a number ）

Time consuming versions

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

A short time-consuming version

class Solution {
public:
    bool isPalindrome(int x) {
        if(x<0)
            return false;
        int len = 1;
        bool temp=true;
        while(x / len >= 10) { //len Of 1 The location of is current x The position of the first digit of 
            len = len * 10;
        }
        while(x){
            int left = x/len;
            int right = x%10;
            if(left!=right){
                temp=false;
                break;
            }
            x = x%len;
            x = x/10;
            len=len/100;  // because x Two missing 
        }
        return temp;
    }
};

（2）python

String version

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if(x<0):
            return False
        s = str(x)
        n = len(s)
        for i in range(n//2):
            j=-(i+1)
            if(s[i]!=s[j]):
                return False
        return True

Non string version

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if(x<0):
            return False
        l = 1
        while(x//l >=10):
            l=l*10
        while(x):
            left = x // l
            right = x % 10
            if(left!=right):
                return False
            x = x%l
            x = x//10
            l = l//100
        return True