String to Integer (atoi)

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.

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.

提交次数最多的一题，总是会溢出。

C++代码如下：

#include<iostream>
#include<cctype>
#include<climits>
using namespace std;

class Solution
{
public:
    int atoi(const char *str)
    {
        const char *p=str;
        int arr[100]= {0};
        int dot=0;
        char flag='+';
        int i=0;
        while(isspace(*p))
            p++;
        //注意符号的判断
        if(*p=='-')
        {
            flag='-';
            p++;
        }
        else if(*p=='+')
            p++;
        //数字的开始位置
        const char *start=p;
        //注意字符数组结束位置的判断
        while(*p!='\0')
        {
            //只能有一个.，且一遇到就返回，因为只取整数
            if(*p=='.')
            {
                if(dot)
                    return 0;
                dot++;
                break;
            }
            //只有刚开始的地方不是数字才会错误，取完数字之后后面出现字符不会报错，例如134aa，返回134
            if(p==start&&!isdigit(*p))
                return 0;
            if(isdigit(*p))
            {
                arr[i]=*p-'0';
                p++;
                i++;
            }
            else
                break;
        }
        long long num=1;
        dot=p-start;
        while(--dot>0)
            num*=10;
        long long sum=0;
        i=0;
        while(p!=start)
        {
            sum+=arr[i]*num;
            //在计算每一次都判断sum，不能等计算完成之后判断，不然会溢出
            if(flag=='+'&&sum>INT_MAX)
                return INT_MAX;
            if(flag=='-'&&-sum<INT_MIN)
                return INT_MIN;
            i++;
            num/=10;
            p--;
        }
        if(flag=='+')
            return sum;
        else
            return -sum;
    }
};

int main()
{
    const char *str="  9223372036854775809";
    Solution s;
    cout<<s.atoi(str)<<endl;
}

运行结果：

 

需要考虑的问题是：1）要删除空格 2）要判断正负号 3）要判断溢出问题。。不需要考虑是否是浮点数。。。。。

 

方法二：

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

class Solution {
public:
    int atoi(string str) {
        if(str.empty())
            return 0;
        int i=0;
        while(str[i]==' ') i++;
        int flag=1;
        if(str[i]=='-')
        {
            flag=-1;
            i++;
        }
        else if(str[i]=='+')
            i++;
        long long res=0;
        while(str[i]!='\0'&&isdigit(str[i]))
        {
            res=res*10+(int)(str[i]-'0');
            if(flag==1&&res>INT_MAX)
                return INT_MAX;
            else if(flag==-1&&-res<INT_MIN)
                return INT_MIN;
            i++;
        }
        return flag*res;
    }
};

int main()
{
    const char *str="    775809";
    Solution s;
    cout<<s.atoi(str)<<endl;
}

 

转载于:https://www.cnblogs.com/wuchanming/p/4098998.html