[LeetCode] String to Integer (atoi) 字符串

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.

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.

 

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　　题目是实现string 转换int 的功能，需要注意的是前置空格，但是会忽略后置部分，需要注意的是判断int 的溢出，可以通过这样的一个判断：

if( res> (INT_MAX - str[i] +'') / )
    overflow = ;
end;

 

#include<string>
#include<iostream>
using namespace std;

class Solution {
public:
    int atoi(string str) {
        if(str.length()==) return ;
        int len = str.length();
        int idx= ;
        while(idx<len&&str[idx]==' ')    idx++;
        if(idx==len)    return ;
        int flag =;
        if(str[idx]=='+')   idx++;
        else if(str[idx]=='-'){
            flag=;
            idx++;
        }
        int sum = ;
        int overflow =;
        while(idx<len&&str[idx]>=''&&str[idx]<=''){
            if(sum>(INT_MAX - str[idx]+'')/){
               overflow = ;
                break;
            }
            sum *=;
            sum+=str[idx] -'';
            idx++;
        }
        if(overflow){
            if(flag)    return INT_MAX;
            else    return INT_MIN;
        }
//        while(idx<len&&str[idx]==' ')   idx++;
//        if(idx!=len)    return 0;
        if(flag)    return sum;
        else    return -sum;
    }
};

int main()
{
    string str = "  2147483648  ";
    Solution sol;
    cout<<sol.atoi(str)<<endl;
    return ;
}