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leetcode8. String to Integer

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問題描述:

Implement atoi which converts a string to an integer.

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.

Note:

  • Only the space character ‘ ‘ is considered as whitespace character.
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [?231, 231 ? 1]. If the numerical value is out of the range of representable values, INT_MAX (231 ? 1) or INT_MIN (?231) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is ‘-‘, which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit ‘3‘ as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is ‘w‘, which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (?231) is returned.

思路:

根據題目的思路來一步一步的做:

首先先把空格都刪除,遇到第一個非空格字符,判斷是否是符號‘+’或‘-’來確定數字的符號,若第一個非空格字符串不為符號也不為數字那麽就返回0。

若為有效字符需要判斷數字的大小。

代碼:

class Solution {
public:
    int myAtoi(string str) {
        if(str.empty())
            return 0;
        int SLen = str.size();
        int ret = 0;
        int sign = 1;
        int i = 0;
        while(i < SLen && str[i] ==  ){
            i++;
        }
        if(str[i] == - || str[i] == +){
            sign = str[i] == - ? -1 : 1;
            i++;
        }
        while(i < SLen && str[i] - 0 > -1 && str[i] - 0 < 10){
            if(ret > INT_MAX / 10 || (ret == INT_MAX / 10 && str[i] - 0 > 7))
                return sign == 1 ? INT_MAX : INT_MIN;
            ret = ret*10 + (str[i] - 0);
            i++;
        }
        return sign * ret;
    }
};

需要註意的點:

1. str中的為字符串,一定要經過str[i] - ‘0‘之後才可以當作數字操作:

(ret == INT_MAX / 10 && str[i] - ‘0‘ > 7)

後面的一段判斷句中又犯過這種粗心的錯誤。

2. 前一段判斷中要確定ret到底與誰相比較,是INT_MAX/10

leetcode8. String to Integer