LeetCode9-Palindrome Number(C++)

Description

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121 </br>
Output: true </br>

Example 2:

Input: -121 </br>
Output: false </br>
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10 </br>
Output: false </br>
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Coud you solve it without converting the integer to a string?

AC代码

class Solution {
public:
    bool isPalindrome(int x) {
        if(x < 0) return false;
        int y = 0, z = x;
        short int n=0;
        while(z != 0) {
            n = z % 10;
            // 这一步是因为要确保 y*10 +n < INT_MAX
            if(y > ((INT_MAX - n)/10)) return false;
            y = y*10 + n;
            z = z/10;
        }
        return (x==y);
    }
    // 这个函数仅仅是用来测试方便，提交的时候也可以不加
    void judgeResult(int x) {
        if(isPalindrome(x)) {
            cout << x << " is palindrome" <<endl;
        } else {
            cout << x << " is not palindrome" <<endl;
        }
    }
};

测试代码

int main() {
    Solution s;
    s.judgeResult(-121);
    s.judgeResult(121);
    s.judgeResult(456);
    s.judgeResult(567898765);
}

测试结果

-121 is not palindrome
121 is palindrome
456 is not palindrome
567898765 is palindrome

总结

前面有类似的题目，是判断字符串是否是回文字符串，这道题的描述里面说了在解题时应避免先转换成字符串再做回文。以上的AC代码实际上是利用了之前某一道题目里面的反转字符串，然后判断字符串反转后是否还与本身相等。