public class Solution { public int MyAtoi(string str) { int index = 0, sign = 1, total = 0; //1. Empty string if (str.Length == 0) { return 0; } //2. Remove Spaces while (str[index] == ' ' && index < str.Length) { index++; } //3. Handle signs if (str[index] == '+' || str[index] == '-') { sign = str[index] == '+' ? 1 : -1; index++; } //4. Convert number and avoid overflow while (index < str.Length) { int digit = str[index] - '0'; if (digit < 0 || digit > 9) break; //check if total will be overflow after 10 times and add digit if (int.MaxValue / 10 < total || int.MaxValue / 10 == total && int.MaxValue % 10 < digit) { return sign == 1 ? int.MaxValue : int.MinValue; } total = 10 * total + digit; index++; } return total * sign; } }
https://leetcode.com/problems/string-to-integer-atoi/#/description
补充一个python的实现：
1 class Solution: 2 def myAtoi(self, string: str) -> int: 3 string = string.strip() 4 n = len(string) 5 if n == 0: 6 return 0 7 sign = 0 8 convertStr = '0' 9 firstNum = False 10 for i in range(n): 11 c = ord(string[i]) - ord('0') 12 if not firstNum: 13 if string[i] == '+' and sign == 0: 14 sign = 1 15 elif string[i] == '-' and sign == 0: 16 sign = -1 17 elif c >= 0 and c <= 9: 18 firstNum = True 19 if sign == 0: 20 sign = 1 21 convertStr += str(c) 22 else: 23 convertStr = '0' 24 break 25 else: 26 if c >= 0 and c <= 9: 27 convertStr += str(c) 28 else: 29 break 30 r = int(convertStr) * sign 31 if r > 2 ** 31 - 1: 32 r = 2 ** 31 - 1 33 elif r < -(2 ** 31): 34 r = -(2 ** 31) 35 return r