description:

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.

spoilers alert… click to show requirements for atoi.

my solution:

int myAtoi(char* str) {
  long long sum = 0;
  bool flag = true;
  bool track = true;
  bool track2 = true;
  for (int i = 0; str[i] != '\0'; i++) {
    if ((str[i] == '+' || str[i] == '-') && track) { track = false; track2 = false; if (str[i] == '-') flag = false; continue; }
    else if (str[i] == '+' || str[i] == '-') return 0;
    if (str[i] >= '0' && str[i] <= '9' && flag) { 
      sum += str[i] - '0';
      if (sum > 2147483647) { return 2147483647; }
      track2 = false;
    }
    else if (str[i] >= '0' && str[i] <= '9' && !flag) { 
      sum -= str[i] - '0';
      if (sum < -2147483648) { return -2147483648; }
      track2 = false;
    }
    else if (str[i] == ' ' && track2) continue;
    else return sum;
    if (str[i + 1] != '\0' && str[i + 1] >= '0' && str[i +1] <= '9') sum *= 10;
  }
  return sum;
}

better ways:

by yuruofeifei 

int atoi(const char *str) {
    int sign = 1, base = 0, i = 0;
    while (str[i] == ' ') { i++; }
    if (str[i] == '-' || str[i] == '+') {
        sign = 1 - 2 * (str[i++] == '-'); 
    }
    while (str[i] >= '0' && str[i] <= '9') {
        if (base >  INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7)) {
            if (sign == 1) return INT_MAX;
            else return INT_MIN;
        }
        base  = 10 * base + (str[i++] - '0');
    }
    return base * sign;
}

thought:

注意溢出的范围，学习了一个快速把字符串转化成整形的方法。