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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> LeetCode 第九題, Palindrome Number

LeetCode 第九題, Palindrome Number

編輯:C++入門知識

LeetCode 第九題, Palindrome Number


題目原文:

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

Some hints:

Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.

題意解析:

即是判斷一個數字是否是回文串.

水題~~~~~~~~~~~~~~~~~~~~~~~~~

代碼如下:

solution 1:依次取出左右的數字,對比.

class Solution {
public:
    bool isPalindrome(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (x < 0)
            return false;
        if (x == 0)
            return true;
            
        int base = 1;
        while(x / base >= 10)
            base *= 10;
            
        while(x)
        {
            int leftDigit = x / base;
            int rightDigit = x % 10;
            if (leftDigit != rightDigit)
                return false;
            
            x -= base * leftDigit;
            base /= 100;
            x /= 10;
        }
        
        return true;
    }
};
solution 2: 將數字取反....判斷是否相等,不過,翻轉的時候可能會溢出,但是,也是過了..............

class Solution {
public:
    bool isPalindrome(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(x < 0) return false;
        
        long long target = x;
        //if(target < 0) target = -target; 
        long long reverse = 0;
        long long curNum = target;
        while(curNum != 0)
        {
            reverse = 10*reverse+curNum%10;
            curNum /= 10;
        }
        
        return reverse == target;
    }
};


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