//此题是easy题，比较简单，主要困难在考虑全输入的各种情况： //1、开始的时候有空格等空白字符 //2、开头有加减号 //3、溢出（第一次写就是没有考虑到这个情况） //C代码 int myAtoi(char* str) {    int i=0;    double result = 0;    int IsNegative = 0;        while(isspace(str[i]))    {        i++;    }        if(str[i] == '-')    {        IsNegative = 1;        i++;    }    else if(str[i] == '+')    {        IsNegative = 0;        i++;    }    else    {            }      for(;i
     
       INT_MAX)    {        result = INT_MAX;    }        if(result < INT_MIN)    {        result = INT_MIN;    }        return (int)result;    }
     

 

#python代码 class Solution(object):    def myAtoi(self, str):        """        :type str: str        :rtype: int        """        num = "1234567890"        INT_MAX = 2147483647        INT_MIN = -2147483648        i = 0        sum = 0        flag = 1                str = str.strip()        if len(str) == 0:            return 0        else:            if str[i] == '-':                flag = -1                i+= 1            elif str[i] == '+':                flag = 1                i+= 1                            if i >= len(str):                return 0                            while i
     
      = sum:                    sum = sum*10                else:                    if flag == 1:                        return INT_MAX                    if flag == -1:                        return INT_MIN                                        if INT_MAX - tempint >= sum:                    sum += tempint                else:                    if flag == 1:                        return INT_MAX                    if flag == -1:                        return INT_MIN                                        i+=1                            return sum*flag