关于c:8字符串转换整数LeetCodeC语言

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/*
** 将字符串转换为数字——LeetCode 第 8 题
** author: aliao
*/
#include <stdio.h>
#include <limits.h>

int myAtoi(char *str)
{
  int i = 0;
  int start = 0;
  int len = 0;
  // 遍历字符串中起始的空格字符
  while (str[i] == 32)
  {i++;}
  // 如果非空首字符不是 -+ 或者 0 - 9 这三类字符，则间接返回 0
  // 如果是空字符串或者字符串只蕴含空白子字符
  if (str[i] != '-' && str[i] != '+' && (str[i] < 48 || str[i] > 57))
  {return 0;}
  else
  {
    start = i;
    len += 1;
    i++;
  }
  while (str[i] != '\0')
  {if (str[i] >= 48 && str[i] <= 57)
    {
      len += 1;
      i++;
    }
    else
    {break;}
  }
  int reverse = 0;
  int k = 10;
  int isNeg = 0;
  for (int j = 0; j < len; j++)
  {if (j == 0)
    {if (str[start + j] == '-')
      {
        isNeg = 1;
        continue;
      }
      else if (str[start + j] == '+')
      {continue;}
    }
    if (isNeg)
    {if (reverse < INT_MIN / 10 || reverse == INT_MIN / 10 && str[start + j] - 48 > 8)
      {return INT_MIN;}
      else
      {reverse = reverse * k - (str[start + j] - 48);
      }
    }
    else
    {if (reverse > INT_MAX / 10 || reverse == INT_MAX / 10 && str[start + j] - 48 > 7)
      {return INT_MAX;}
      else
      {reverse = reverse * k + (str[start + j] - 48);
      }
    }
  }
  return reverse;
}
int main (void) {
  char *str = "-2147483648";
  printf("final value: %d", myAtoi(str));
}