8. String to Integer (atoi)

   

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• Difficulty: Medium
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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.

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 //1. 若字符串开头是空格，则跳过所有空格，到第一个非空格字符，如果没有，则返回0. 2  3 //2. 若第一个非空格字符是符号+/-，则标记sign的真假，这道题还有个局限性，那就是在c++里面，+-1和-+1都是认可的，都是-1，而在此题里，则会返回0. 4  5 //3. 若下一个字符不是数字，则返回0. 完全不考虑小数点和自然数的情况，不过这样也好，起码省事了不少。 6  7 //4. 如果下一个字符是数字，则转为整形存下来，若接下来再有非数字出现，则返回目前的结果。 8  9 //5. 还需要考虑边界问题，如果超过了整形数的范围，则用边界值替代当前值。10 11 class Solution {12 public:13     int myAtoi(string str) {14         if (str.empty()) return 0;//判断string的empty15         int i = 0;16         int sign = 1;17         int base = 0;18         int n = str.size();19         while(i < n && str[i] == ' '){
    //跳过空格20             i++;21         }22         23         if (str[i] == '+' || str[i] == '-') {
    //判断正负24             sign = str[i++] == '+' ? 1 : -1;25         }26         //sign = str[i++] == '+' ? 1 : -1;27         28         //for(;i < n; i++){
     29         while(i < n && str[i] >= '0' && str[i] <= '9'){
    //别忘了还有一个条件：i
      
        INT_MAX / 10 ||(base == INT_MAX / 10 && str[i] - '0' > 7)){
    //要先判断是否溢出31                 return sign == 1 ? INT_MAX : INT_MIN;//-2147483648 ~ 214748364732             }33             base = base * 10 + (str[i++] - '0') ;34         }35             36         return sign * base;37     }38 };