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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.
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.
1 int myAtoi(char* str) { 2 int sign = 1; //设置符号位,初始设置1为正数 如果前面没有符号就代表正数 3 long long sum = 0; //sum范围要大于int 4 if(str == NULL) 5 return 0; 6 while(*str == ‘ ‘) str++; //去掉前面空格 7 if(*str == ‘-‘){ 8 sign = -1; 9 str++; 10 } //取符号 11 else if(*str == ‘+‘){ 12 sign = 1; 13 str++; 14 } 15 while(*str != ‘\0‘){ 16 if(*str < ‘0‘ || *str > ‘9‘) //不符合规则的取消，如“ +-123” 17 break; 18 else if(*str >= ‘0‘ && *str <= ‘9‘){ 19 sum = sum * 10 + *str -48; 20 if(sum > 2147483648) //sum范围超出 21 break; 22 } 23 str++; 24 } 25 if(sign == 1 && sum > 2147483647) 26 return 2147483647; 27 if(sign == -1 && sum > 2147483648) 28 return -2147483648; 29 sum = (int)sum * sign; //最后把sum转化成int类型 30 return sum; 31 }
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原文地址：http://www.cnblogs.com/boluo007/p/5470627.html
