LeetCode 8 String to Integer (atoi)

8. String to Integer (atoi)

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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.

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实现atoi函数

字符串转整数.

需要注意一下几点

1. 字串为空或者全是空格，返回0； 

2. 字串的前缀空格需要忽略掉；

3. 忽略掉前缀空格后，遇到的第一个字符，如果是‘+’或‘－’号，继续往后读；如果是数字，则开始处理数字；如果不是前面的2种，返回0；

4. 处理数字的过程中，如果之后的字符非数字，就停止转换，返回当前值；

5. 在上述处理过程中，如果转换出的值超出了int型的范围，就返回int的最大值或最小值。

class Solution {public:    int myAtoi(string str)    {        long long ret=0;        int len=str.length();        if(len==1&&str[0]>='0'&&str[0]<='9')                return str[0]-'0';        int pos=0;        while(pos<len&&str[pos]==' '){pos++;}        if(pos==len)                return 0;        int flag=1;        if(str[pos]=='+')                flag=1;        else if(str[pos]=='-')                flag=-1;        else if(str[pos]>'9'||str[pos]<'0')                flag=0;        else            pos--;        for(int i=pos+1;i<len&&str[i]>='0'&&str[i]<='9';i++)        {                ret=10*ret+(str[i]-'0');                if(ret*flag>=INT_MAX)                {                        ret=INT_MAX;                        break;                }                if(ret*flag<=INT_MIN)                {                        ret=INT_MIN;                        break;                }        }        ret*=flag;        return (int)ret;    }};