• 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.

    • 字符串简单题,主要是注意细节,看定义里时如何处理输入字符串的。判断不规则但是有效输入,无效格式,溢出数据。处理溢出数据时,要注意判断条件。
    • 判断溢出时需要注意 abs(INT_MIN) (-2147483648) > abs(INT_MAX) (2147483647)。
    • 两种解法思想一样,就是第一种用了isdigit()判断数字,更简洁。
    • 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 }
    View Code
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  • 原文地址:https://www.cnblogs.com/pegasus923/p/7449544.html
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