• LeetCode OJ--Palindrome Number


    https://oj.leetcode.com/problems/palindrome-number/

    判断是否为回文数

    取每一位存到vector中,再判断

    负数不是回文数

    class Solution {
    public:
        bool isPalindrome(int x) {
            vector<int> digit;
            if(x<0)
                return false;
    
            while(x)
            {
                digit.push_back(x%10);
                x = x/10;
            }
    
            for(int i = 0;i<digit.size()/2;i++)
            {
                if(digit[i]!=digit[digit.size() - 1 -i])
                    return false;
            }
            return true;
        }
    };
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  • 原文地址:https://www.cnblogs.com/qingcheng/p/3822561.html
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