算法设计与分析课作业【week6】leetcode--8. String to Integer (atoi)

题目

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: [−2^31,  2^31 − 1]. If the numerical value is out of the range of representable values, INT_MAX (2^31 − 1) or INT_MIN (−2^31) 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.

题目的要求是从一个字符串中读取数值，从头开始，碰到第一个非空字符时读取数值，读到非数字时停止（第一个字符正负符号除外）。

那么我们可以首先除去字符串刚开始的空字符，再来读取字符，这里我们需要注意以下情况：

1. 如果遇到'+'、'-'符号，可以读取，但要注意只能读取一次。如"+-1"，这个是错误的，而且只能在第一个读取，如果出现"1544+54"，在读到'+'号就需要停止了。 
2. 遇到非数字则停止读取。
3. 根据题目要求，在读取数字时需要注意数字的大小，超过INT_MAX或者小于INT_MIX则可以停止读取了。

C++代码如下：

class Solution {
public:
    int myAtoi(string str) {
        /* negative用来记录数值正负，
           start用来判断确定已经开始读取数值，避免"12+13"的读取错误情况。*/
        bool negetive = false, start = false;
        int result = 0, i = 0;
        
        //首先除去空白字符
        while (str[i] == ' ') 
            ++i;
       
        for ( ; i < str.length(); ++i) {
            if (!start && (str[i] == '-' || str[i] == '+')) {
                negetive = str[i] == '-' ? true : false;
                start = true;
            }
            else if (str[i] <= '9' && str[i] >= '0') {
                start = true;
                
                //判断数值大小可否超过限定最大最小数值
                if (result > INT_MAX / 10|| result == INT_MAX / 10 && (str[i] - 48) > INT_MAX % 10)
                    return negetive ? INT_MIN : INT_MAX;
                result = result * 10 + str[i] - 48; 
            }
            else 
                break;
        }
        return negetive ? -result : result;
    }
};