LeetCode刷题记录3-求数字是否为回文

求一个数字是否为回文

[2017-11-23修改：增加字符串回文判断]

先写一个本人的笨办法，不过可以判断负数，该方法把整个数字每一位都存了下来(其实可以只存一半)：

bool isPalindrome(int x){    int temp = x;    vector<int> vec;    while (temp)    {        vec.push_back(temp % 10);        temp /= 10;    }    std::reverse(vec.begin(), vec.end()); //从后往前比较    //只需要比较一半即可    for(int i = 0;  i < vec.size() / 2 + 1; ++i)    {        if (x % 10 != vec[i])        {            return false;        }        x /= 10;    }    return true;}

接下来是LeetCode的解法：

bool IsPalindrome(int x) {        // As discussed above, when x < 0, x is not a palindrome.        // Also if the last digit of the number is 0, in order to be a palindrome,         // the first digit of the number also needs to be 0.        // Only 0 satisfy this property.        if(x < 0 || (x % 10 == 0 &&        if(x < 0 || (x % 10 == 0 && x != 0)) {            return false;        }        int revertedNumber = 0;        while(x > revertedNumber) {            revertedNumber = revertedNumber * 10 + x % 10;            x /= 10;        }        // When the length is an odd number, we can get rid of the middle digit by revertedNumber/10        // For example when the input is 12321, at the end of the while loop we get x = 12, revertedNumber = 123,         // since the middle digit doesn''t matter in palidrome(it will always equal to itself), we can simply get rid of it.        return x == revertedNumber || x == revertedNumber/10;    }

判断回文字符串

提供一个判断回文字符串的方法：

bool isPalindrome__(const std::string& in){    std::string res = std::string(in.begin(),        std::mismatch(in.begin(), in.end(), in.rbegin()).first);    return in == res;}

mismatch的实际算法源码如下，返回一个pair包括2个容器迭代器的位置：

template<class InputIt1, class InputIt2>std::pair<InputIt1, InputIt2>    mismatch(InputIt1 first1, InputIt1 last1, InputIt2 first2){    while (first1 != last1 && *first1 == *first2) {        ++first1, ++first2;    }    return std::make_pair(first1, first2);}