LeetCode#8. String to Integer (atoi)

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.

问题大意：将字符串转为对应的数字。该字符串只有数字，正负号，和开头的空格组成（如果有的话）。

解决思路：遍历字符串，找到第一个不为空格的字符，判断其是否为数字（isdigital）,如果是就转为数字；如果不是再看其是否为“+”或者“-”，将符号置为+1或者-1；最后如果不是前两者，终止遍历。

#include<string>
#include<iostream>
using namespace std;

int myAtoi(string s) {
	long long ans = 0;
	int sign = 0, i = 0;
	for (; i < s.size(); i++)
		if (s[i] != ' ')
			break;

	for (; i < s.size(); i++) {
		if (isdigit(s[i])) {
			ans = ans * 10 + s[i] - '0';//最重要就这一句
			if (sign != -1 && ans >= INT_MAX) 
				return INT_MAX;
			else if (sign == -1 && -ans <= INT_MIN) 
				return INT_MIN;
		}
		else if (s[i] == '+' || s[i] == '-') {
			if (sign) break;//碰到第二个+-号就当做终止标志
			sign = s[i] == '-' ? -1 : 1;
		}
		else //除了数字和+-号，其他所有字符都作为结束标志
			break;
	}
	return sign != 0 ? ans * sign : ans;
}

int main(){
	string s1 = "    -91283472332";
	long long result = myAtoi(s1);
	cout << "字符串转整数输出是： " << result << endl;
	//-2147483648
	getchar();
}

LeetCode#8. String to Integer (atoi)

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