8. String to Integer (atoi) --重要

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.
以下代码仅仅考虑了空格的问题，以及溢出的问题，没有考虑不合法输出的问题，以及进制的问题

int myAtoi(string str)     {        long result = 0;        int indicator = 1;        for(int i = 0; i<str.size();)        {            i = str.find_first_not_of(' '); //第一个i的位置是第一个非空的位置            if(str[i] == '-' || str[i] == '+')                indicator = (str[i++] == '-')? -1 : 1; //正负号的处理，即正负数的处理            while('0'<= str[i] && str[i] <= '9')             {                result = result*10 + (str[i++]-'0');                if(result*indicator >= INT_MAX) return INT_MAX; //判断是否溢出                if(result*indicator <= INT_MIN) return INT_MIN;                            }            return result*indicator;        }    }

int myAtoi(string str) {          int index=0;          while(str[index]==' ')index++;          int flag=1;          if(str[index]=='-'){              index++;              flag=-1;          }else if(str[index]=='+'){              index++;          }          int res=0;          while(index<str.size()){              if(str[index]<'0' || str[index]>'9'){                  return flag*res;              }              int digit=str[index]-'0';              if(flag==1 && res*10.0+digit>INT_MAX){                  return INT_MAX;              }else if(flag==-1 && -(res*10.0+digit)<INT_MIN){                  return INT_MIN;              }              res = res*10+digit;              index++;          }          return flag*res;      } 

剑指offer中则考虑的全面些，考虑全面的代码如下：

enum status  {kvalid=0,kinvalid}; //该枚举型变量用来标示如果返回num=0的情况下，字符串是合法输入还是非法输入，初始值kvalid =0(合法输入)//kinvalid =1（非法输入），合法非法的状态存储在全局变量g_status中 int g_status = kvalid; int strToIntCore(string str, int start, int end, bool minus){    int flag = minus?-1:1;    long long num = 0;    while(str[start] != '\0' && start<=end){        if(str[start] >= '0' && str[start] <= '9'){            num = num * 10 + flag * (str[start] - '0');            if(num > INT_MAx || num < INT_MIN){ //判断数字是否溢出 ，溢出至num=0                 num = 0;                break;            }            start++;        }        else{ //不合法输入             num = 0;            break;        }    }    if(str[start] == '\0'){        g_status = kvalid;    }     return num;}int strToInt(string str){    int g_status = kinvalid;    int len = str.length();    int start = 0;    int end = len - 1;    long long num = 0;    bool minus = false;//标示正负数     if(len >= 1){        while(str[start] == ' ') //处理一开始的空格 ，诸如“    +4500”             start++;        if(str[start] == '+'){             start++;        }        else if(str[start] == '-'){            start++;            minus = true;        }        if(start <= end){            num = strToIntCore(str,start,end,minus);        }    }       return num;}

其实还可以考虑进制的因素：设置一个radius记录进制
num = num * radius + str[i] - ‘0’,即可；
其中16进制的数字开头为“0X”
8进制的数字开头为”0”
剩下设置为十进制的数字