LeetCode 8. String to Integer (atoi)

https://leetcode.com/problems/string-to-integer-atoi/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.

Requirements for atoi:

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

• 字符串簡單題，主要是注意細節，看定義裏時如何處理輸入字符串的。判斷不規則可是有效輸入，無效格式，溢出數據。處理溢出數據時，要注意判斷條件。
• 判斷溢出時須要注意 abs(INT_MIN) (-2147483648) > abs(INT_MAX) (2147483647)。
• 兩種解法思想同樣，就是第一種用了isdigit()判斷數字，更簡潔。
• atoi - C++ Reference
  • http://www.cplusplus.com/reference/cstdlib/atoi/
• isdigit - C++ Reference
  • http://www.cplusplus.com/reference/cctype/isdigit/

  1 //
  2 //  main.cpp
  3 //  LeetCode
  4 //
  5 //  Created by Hao on 2017/3/16.
  6 //  Copyright © 2017年 Hao. All rights reserved.
  7 //
  8 
  9 #include <iostream>
 10 #include <cstring>
 11 #include <vector>
 12 using namespace std;
 13 
 14 class Solution {
 15 public:
 16     int myAtoi(string str) {
 17         int i = 0;
 18         int result = 0;
 19         
 20         if (str.empty()) return 0;
 21         
 22         while (' ' == str.at(i) && i < str.size()) ++ i;
 23         
 24         if (i == str.size()) return 0;
 25         
 26         int sign = 1;
 27         
 28         if ('+' == str.at(i)) {
 29             sign = 1;
 30             ++ i;
 31         } else if ('-' == str.at(i)) {
 32             sign = -1;
 33             ++ i;
 34         }
 35         
 36         while (i < str.size() && isdigit(str.at(i))) {       // isdigit() from <cctype>
 37             if (result > (INT_MAX - (str.at(i) - '0')) / 10) // use > instead of >= for that abs(INT_MIN) (-2147483648) > abs(INT_MAX) (2147483647)
 38                 return -1 == sign ? INT_MIN : INT_MAX;
 39             
 40             result = result * 10 + (str.at(i) - '0');
 41             
 42             ++ i;
 43         }
 44         
 45         return result * sign;
 46     }
 47     
 48     int myAtoi2(string str) {
 49         int num = 0;
 50         int sign = 1;
 51         int i = 0;
 52         const int n = str.length();
 53         
 54         if (n == 0) return 0;
 55         
 56         while ((str[i] == ' ') && (i < n))
 57             i ++;
 58         
 59         if (str[i] == '+')
 60             i ++;
 61         else if (str[i] == '-') {
 62             sign = -1;
 63             i ++;
 64         }
 65         
 66         while (i < n) {
 67             if (str[i] < '0' || str[i] > '9')
 68                 break;
 69             
 70             // Be careful for out of range value
 71             if ((num > INT_MAX / 10) || ((num == INT_MAX / 10) && ((str[i] - '0') > INT_MAX % 10)))
 72                 return sign == -1 ? INT_MIN : INT_MAX;
 73             
 74             num = num * 10 + (str[i] - '0');
 75             
 76             i ++;
 77         }
 78         
 79         return num * sign;
 80     }
 81 };
 82 
 83 int main(int argc, char* argv[])
 84 {
 85     Solution    testSolution;
 86     string      result;
 87     
 88     vector<string> sVec = {"      -11919730356x", "", "1", "-2147483648", "-2147483647", "2147483647", "2147483648", "-3924x8fc", "+ 413", "++c", " ++1", "+413"};
 89     
 90     /*
 91      -2147483648
 92      0
 93      1
 94      -2147483648
 95      -2147483647
 96      2147483647
 97      2147483647
 98      -3924
 99      0
100      0
101      0
102      413
103      */
104     for (auto s : sVec) {
105         cout << testSolution.myAtoi(s) << endl;
106         cout << testSolution.myAtoi2(s) << endl;
107     }
108 
109     return 0;
110 }

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