【LeetCode】LeetCode——第8题：String to Integer (atoi)

8. String to Integer (atoi)

    My Submissions

Question Editorial Solution

Total Accepted: 98161  Total Submissions: 728890  Difficulty: Easy

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

spoilers alert... click to show requirements for atoi.

Requirements for atoi:

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

Subscribe to see which companies asked this question

Show Tags

Show Similar Problems

题目的大概意思是：按照atoi函数的转换原则，将输入的字符串转换成整型数。其实就是自行实现atoi函数。

这道题难度等级：简单

需要注意的是以下几点：

但是怎么觉得不止这几种情况啊，这些情况我都考虑且避免了。还是AC不掉，暂时没有找到一个详细的转换原则。

网上找到的代码如下：

class Solution {
public:
    int myAtoi(string str) {
        int length = str.size();
        long long ret_64 = 0;
        int op = 1;
        int p = 0;
        while (str[p] == ' ') ++p;
        if (str[p] == '+' || str[p] == '-') {
            if (str[p] == '-') op = -1;
            p++;
        }

        for (int i = p; i < length; ++i)
            if ('0' <= str[i] && str[i] <= '9') {
                ret_64 = ret_64 * 10 + (str[i] - '0');
                if ((op == -1 && ret_64 > 2147483648LL)) return -2147483648;
                if ((op == 1 && ret_64 > 2147483647LL)) return 2147483647;
            }   else break;
        return (int)ret_64 * op;
    }
};

提交后顺利AC掉该题，Runtime: 12 ms。