字符串转成整型(int)

Posted 2020-10-30 zhchoutai

1 题目

Implement atoito convert a string to an integer.

Hint: Carefullyconsider all possible input cases. If you want a challenge, please do not seebelow and ask yourself what are the possible input cases.

Notes: It isintended 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 functionfirst discards as many whitespace characters as necessary until the firstnon-whitespace character is found. Then, starting from this character, takes anoptional initial plus or minus sign followed by as many numerical digits aspossible, and interprets them as a numerical value.

The string cancontain additional characters after those that form the integral number, whichare ignored and have no effect on the behavior of this function.

If the firstsequence of non-whitespace characters in str is not a valid integral number, orif no such sequence exists because either str is empty or it contains onlywhitespace characters, no conversion is performed.

If no validconversion could be performed, a zero value is returned. If the correct valueis out of the range of representable values, INT_MAX (2147483647) or INT_MIN(-2147483648) is returned.

2 分析

(1) 空字符串返回0；

(2) 去除字符串前边的0；

(3) 若遇到非数字字符则终止计算；

(4) 若求出的数越界则返回最大值或者最小值。

3 实现

int atoi(string str)
{
	int start = str.find_first_not_of(‘ ‘);
	if (start > 0)
	{
		str = str.substr(start);
	}
	int size = str.size();
	if (0 == size)
	{
		return 0;
	}

	int signal = 1;
	int i = 0;
	if (str[i] == ‘+‘)
	{
		++i;
	}
	else if (str[i] == ‘-‘)
	{
		signal = -1;
		++i;
	}

	long long res = 0;
	while (i < size)
	{
		if (str[i] < ‘0‘ || str[i] > ‘9‘)
		{
			break;
		}
		res = res * 10 + (str[i] - ‘0‘);
		if (res > INT_MAX)
		{
			return signal == 1 ? INT_MAX : INT_MIN;
		}
		++i;
	}

	return res * signal;
}