[Leetcode] String to Interger (atoi)

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本文介绍了一个atoi函数的实现方法，该函数用于将字符串转换为整数。文章详细解释了如何处理输入字符串中的空白字符、正负号及数字字符，并确保转换结果在整数范围内。

String to Integer (atoi) Dec 27 '11 6739 / 30948

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.

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.

» Solve this problem

#define  INT_MAX (2147483647)
#define  INT_MIN (-2147483648)
class Solution {
public:
    int atoi(const char *str) {
        char c;
        int len = strlen(str);
        int num = 0;
        bool neg = false;
        bool start = true;
        int old;
        for (int i =0 ; i < len; i++) {
            c = str[i];
            if (start && isspace(c)) continue;
            if (!start && isspace(c)) break;
            if (start) {
                if (c == '-') { 
                    neg = true;
                    start =false;
                    continue;
                }
                else if (c == '+') {
                    neg = false;
                    start = false;
                    continue;
                }
                start = false;
            }
            if (c < '0' || c > '9') break;
            old = num;
            num = 10 * num + (c - '0');
            if ((num - (c - '0'))/ 10 != old) { 
                if (neg) return INT_MIN;
                else return INT_MAX;
            }
        }
        if (num < 0) {
            if (neg) return INT_MIN;
            else return INT_MAX;
        }
        if (neg) {
            return -num;
        }
        else return num;
    }
};