LeetCode (8) String to Integer (atoi) C语言程序

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这道题的要求如下，要注意的事项还是很多的，需要考虑各种情况。

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.

C语言代码如下，AC。

int myAtoi(char* str) {        if (str == NULL )//空字符串时，返回0.       return 0;         int i;    for (i=0; str[i] != '\0' && str[i] == ' '; i++)       ;    if (str[i] == '\0')//字符串全为空格时，返回0.       return 0;        int sign = 1;// 符号必须初始化为1，因为如果不初始化，只写为 int sign 的话，如果第一个非空字符不是+/-，将会得出错误答案。    if (str[i] == '+'){        i++;    }    else if (str[i] == '-') {        sign = -1;        i++;    }        int MAX = 0x7fffffff;    int MIN = 0x80000000;    int j;    long temp = 0;        for ( j = i; str[i] != '\0' ; j++){                if (str[j] >= '0' && str[j] <= '9'){            temp = temp * 10 + sign * (str[j] - '0');            if (temp > MAX || temp < MIN) ////处理溢出的情况            return (temp > 0) ? MAX : MIN;        }        else           break;// 如果后面的字符不再是数字，那么停止转换，返回现有值.     }     return (int) temp;}

参考博客：http://blog.csdn.net/zhouworld16/article/details/16082765

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