[LeetCode]8.String to Integer (atoi)（自学留存）-优快云博客

本文介绍了一个将字符串转换为整数（atoi）的实现方案。该方案首先去除字符串前导空格，然后读取首个非空格字符，并判断是否为有效数值。文章详细解析了如何处理正负号及数字字符，确保正确转换的同时防止整数溢出。

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.

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.

Personal tips: 这是没有使用long类型的判断溢出，实际上使用long能更好判断溢出，以下为代码：

class Solution {
public:
	int myAtoi(string str) {
		int start = 0;
		for (; start < str.size(); ++start)
		{
			if (isspace(str[start]))		continue;
			else
			{
				if ((str[start] == '+' && (str[start + 1] >= '0'&& str[start + 1] <= '9')) ||
					(str[start] == '-' && (str[start + 1] >= '0'&& str[start + 1] <= '9') ||
					(str[start] >= '0'&& str[start] <= '9'))) break;
				else  return 0;
			}
		}
		int num = 0;
		if (str[start] == '+' || (str[start] >= '0'&& str[start] <= '9'))
		{
			if (str[start] == '+') ++start;
			for (; start < str.size(); ++start)
			{
				if (str[start] >= '0'&& str[start] <= '9')
				{
					cout << (str[start] - '0') << endl;
					if ((INT_MAX - (str[start] - '0')) / 10 < num) return INT_MAX;
					else num = num * 10 + (str[start] - '0');
				}
				else break;
			}
		}
		else
		{
			
			for (start = start + 1; start < str.size(); ++start)
			{
				if (str[start] >= '0'&& str[start] <= '9')
				{
					if ((INT_MIN + (str[start] - '0') )/ 10> num) { return INT_MIN; }
					else { num = num * 10 - (str[start] - '0'); }
				}
				else break;
			}
		}
		return num;
	}
};