LeetCode String to Integer (atoi) 字符串转换整数

String to Integer (atoi)

Implementatoito 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.

spoilers alert... click to show requirements for atoi.

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.

好长的注意地方，因为要考虑的东西实在也挺多的。总结如下：

1 前面空格分隔符号的时候

2 第一个符号位处理+ -

3 遇到非数字字符退出

4 为正数的时候，大于INT_MAX上溢

5 为负数的时候， 小于INT_MIN下溢

6 为空字符串或者空指针的时候

考虑完了就好写程序了：

class Solution {
public:
	int atoi(const char *str) 
	{
		if (str == nullptr) return 0;
		//注意：判断空字符串数组
		if(*str == '\0') return 0;
		long long llnum = 0;
		int sign = 1;

		while (*str == ' ')
		{
			str++;
		}
		if (*str == '-')
		{
			sign = -1;
			str++;
		}
		//注意：不要溜了要判断+号的情况
		else if(*str == '+')
		{
			str++;
		}
		while (isdigit(*str))
		{
			int i = *str - '0';
			llnum = llnum*10 + i;
			if (llnum*sign > INT_MAX)
			{
				llnum = INT_MAX;
				break;
			}
			else if (llnum*sign < INT_MIN)
			{
				llnum = INT_MIN;
				break;
			}
			str++;
		}
		return int(sign*llnum);
	}
};

2014-1-24 update

简洁一点的程序，思路一样，这题没多少变化：

int atoi(const char *str) 
	{
		long long rs = 0;
		while (isspace(*str)) str++;
		
		int sign = 1;
		if (*str == '-')
		{
			sign = -1;
			str++;
		}
		else if (*str == '+') str++;
		
		for ( ; isdigit(*str); str++)
		{
			rs = rs*10 + (*str-'0');
			if (sign == 1 && rs >= INT_MAX) return INT_MAX;
			else if (rs*sign < INT_MIN) return INT_MIN;
		}
		return rs*sign;
	}