github地址：https://github.com/ALXlixiong/offer

题目描述：将一串由数字（可能包含正负号）组成的字符串转化为整数；

注意事项：

（1）判断字符串是否为空；

（2）判断数字是否溢出（正负号）

（3）是否含有其他符号

代码：

#include<stdio.h>

int str_num(char *str)
{
  if(*str == '\0' || NULL == str)
  {
    printf("不合法\n");
    return -1;//不合法
  }
  long long num = 0;
  char first = *str;
  int fuhao = 1;   //记录符号
  if(first == '-')
  {
    str++;
    fuhao = -1;
  }
  else if(first == '+')
  {
    str++;
    fuhao = 1;
  }
  if('\0' == *str)
  {
    printf("不合法\n");
    return -1;
  }
  while(*str != '\0')
  {
    if(*str<='9' && *str >= '0')
    {
      num = num*10 + (*str-'0');
      if((fuhao == 1 && num > 0x7fffffff) || (fuhao == -1 && (fuhao *num)<(signed int)0x80000000))
      {
        printf("溢出:num = %lld\n",num);
        return -2;  //溢出
      }    
    }
    else {
      printf("不合法\n");
      return -1;//不合法
    }
    str++;
  }
  return fuhao*num;

}

void Test()
{

  char *string = "-111111234";
  int res = str_num(string);
  printf("res = %d\n",res);

  char *str1 = "\0";
  str_num(str1);
  
  char *str2 = "+";
  str_num(str2);

  char *str3 = "10000000000000000000000000000000";
  str_num(str3);

  char *str4 = "+12-1627";
  str_num(str4);

}

int main()
{
  Test();
  return 0;
}