leetcode--String to Integer (atoi)

1.题目描述

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.

2.解法分析

只需注意几点即可：

• 字符串之前的空格要去除，零串也要去除
• 空串的状况
• 正负号需要考虑
• 溢出管理
• 字符串之中出现异常字符的处理

对于以上几种情况，只有溢出管理比较麻烦，这里设置一个long long 类型的result，计算过程中一旦发现result超出INTMIN-INTMAX定义的int范围，直接返回，代码如下：

class Solution {
public:
    int atoi(const char *str) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(str==NULL)return -1;
        long long result=0;
        int i=0;
        bool isNegative=false;
        bool isHeader=true;
        while(str[i]!='\0')
        {
            if(isHeader)
            {
                if(str[i]==' '){i++;continue;}
                if(str[i]=='-'){isNegative=true;i++;isHeader=false;continue;}
                else
                    if(str[i]=='+'){isHeader=false;i++;continue;}
            }
            if(str[i]<'0'||str[i]>'9')break;
            else
            {
                isHeader=false;
                result=result*10+str[i]-'0';
                if(-result<INT_MIN&&isNegative)return INT_MIN;
                if(result>INT_MAX&&!isNegative)return INT_MAX;
            }
            i++;
        }

        if(isNegative)result*=-1;
        return result;
    }
};