标签：style blog class code c tar
本文为senlie原创，转载请保留此地址：http://blog.csdn.net/zhengsenlie
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.
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.注意规则
2.注意最小的负数的绝对值比最大的正数的绝对值大1，所以 line 13: (str[i] - ‘0‘) > INT_MAX%10不能取等号
class Solution {
public:
int atoi(const char *str) {
int num = 0, sign = 1, i = 0, n = strlen(str);
while(str[i] == ‘ ‘ && i < n) i++;
if(str[i] == ‘+‘) i++;
if(str[i] == ‘-‘) {i++; sign = -1;}
while(i < n){
if(str[i] < ‘0‘ || str[i] > ‘9‘) break;
if(num > INT_MAX/10 || (num == INT_MAX/10 && (str[i] - ‘0‘) > INT_MAX%10)) return sign > 0 ? INT_MAX : INT_MIN;
num = num * 10 + (str[i++] - ‘0‘);
}
return sign * num;
}
};
Leetcode 数 String to Integer (atoi),布布扣,bubuko.com
Leetcode 数 String to Integer (atoi)
标签：style blog class code c tar
原文地址：http://blog.csdn.net/zhengsenlie/article/details/25836439
