LeetCode String to Integer (atoi)

LeetCode String to Integer (atoi)

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

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.


这题要求我们手动实现由string转换为int的atoi函数。当然有很多细节需要考虑,不过你可以一步一步来,先实现常见的case,然后看看出错样例,再逐步完善自己的算法。

错了几次之后,我看了atoi函数的介绍,上面并没有说明如果数值超过int的表示范围应该怎么处理:

If the converted value would be out of the range of representable values by an int, it causes undefined behavior. Seestrtol for a more robust cross-platform alternative when this is a possibility.

不过strtol函数提到如果超出long表示范围则返回最大值或最小值,所以在实现atoi时我也是这么做的。

算法过程是:首先去除前导空白字符(如空格),如果剩余字符串长度为0或者1,则特殊处理;判断有无符号位;然后对字符依次处理,直到遇到一个非数字字符,break;最后检查是否超出int范围以及符号位。

完整代码如下:

class Solution {
public:
	int myAtoi(string str) {
		int i = 0;
		while (i < str.size() && str[i] == ' ')
			i++;
		string tmp = str.substr(i);
		if (tmp.size() == 0 || (tmp.size() == 1 && (tmp[0] < '0' || tmp[0] > '9')))
			return 0;
		string digits = (tmp[0] == '-' || tmp[0] == '+') ? tmp.substr(1) : tmp;
		double ans = 0;
		for (int i = 0; i < digits.size(); i++)
			if (digits[i] < '0' || digits[i] > '9')
				break;
			else
				ans = ans * 10 + (digits[i] - '0');
		ans = (tmp[0] == '-') ? -ans : ans;
		if (ans > INT_MAX)
			return INT_MAX;
		else if(ans < INT_MIN)
			return INT_MIN;
		else return ans;
	}
};

本代码提交AC,用时8MS,居然击败了75%的CPP用户:-)

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