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.
上次面试时考了这道题，有想过处理空格，有问过123ab是怎么的情况，还有就是越界又是怎么样，但是忘了考虑+-号。越界当时也没写，只是和面试官说，哎，真是太不好了。
这次leetcode上一次通过。
1 class Solution { 2 public: 3 int atoi(const char *str) { 4 if (str == NULL) return 0; 5 long long ans = 0; 6 while (*str != ' ' && *str == ' ') str++; 7 int symbol = 1; 8 if (*str == '-') symbol = -1; 9 if (*str == '-' || *str == '+') str++; 10 11 for (; *str != ' ' && *str >= '0' && *str <= '9'; str++) { 12 ans = ans * 10 + (*str - '0'); 13 } 14 ans *= symbol; 15 if (ans <= INT_MIN) return INT_MIN; 16 if (ans >= INT_MAX) return INT_MAX; 17 return ans; 18 } 19 };
代码比起上次写（第二次）都简洁许多。
1 class Solution { 2 public: 3 int atoi(const char *str) { 4 if (str == NULL) return 0; 5 while (*str != ' ' && *str == ' ') str++; //ignore spaces 6 long long ret = 0; 7 long long symbol = 1; 8 if (*str != ' ' && (*str == '+' || *str == '-')) { 9 if (*str == '-') symbol = -1; 10 str++; 11 } 12 13 while (*str != ' ') { 14 if (*str >= '0' && *str <= '9') { 15 ret = ret * 10 + (*str - '0'); 16 } else { 17 break; 18 } 19 str++; 20 } 21 22 while (*str != ' ' && *str == ' ') str++; //ignore spaces 23 ret *= symbol; 24 if (ret > INT_MAX) ret = INT_MAX; 25 if (ret < INT_MIN) ret = INT_MIN; 26 return ret; 27 } 28 };