leetcode: String to Integer (atoi)

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本文介绍了一种将字符串转换为整数（atoi）的方法，包括处理正负号、忽略前导空格、检查非数字字符及防止整数溢出等细节。提供了详细的C++实现代码。

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.

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.

思路

很简单，把每位的数字相应的加起来就行了。这里的重点就是各种情况怎么处理，不过如果不看下面的requrements，可能会有各种不同的处理方法，也就无所谓正确不正确了。我是第5次提交才满足了上面的所有条件，怪不得这道题的通过率那么低。有一点处理就是溢出的判断，比较麻烦一点。当然如果用long类型或者double类型则是另外一个回事。

class Solution {
public:
    int atoi(const char *str) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		int strLen=strlen(str);
		if(strLen==0) return 0;
		const char *head=str;
		for(int i=0;i<strLen;i++)
		{
			if(str[i]==' ') head++;
			else break;
		}
		int numLen=strlen(head);
		if(numLen==0) return 0;
		int sig=1;
		int EXCEPT_LAST=INT_MAX/10;
		int LAST_DIGIT=INT_MAX-EXCEPT_LAST*10;
		int OVERFLOW_ANSWER=INT_MAX;
		if(head[0]=='-')
		{ 
			EXCEPT_LAST=-(INT_MIN/10);
			LAST_DIGIT=-EXCEPT_LAST*10-INT_MIN;
			OVERFLOW_ANSWER=INT_MIN;
			sig=-1;
			numLen--;
			head++;
		}
		else if(head[0]=='+')
		{ 
			EXCEPT_LAST=INT_MAX/10;
			LAST_DIGIT=INT_MAX-EXCEPT_LAST*10;
			OVERFLOW_ANSWER=INT_MAX;
			sig=1;
			numLen--;
			head++;
		}
		else if(head[0]<'0' || head[0]>'9')
		{
			return 0;
		}
		int res=0;
		for(int i=0;i<numLen;i++)
		{
			if(head[i]<'0' || head[i]>'9') break;
			
			if(res<EXCEPT_LAST)
				res=res*10+int(head[i]-'0');
			else if(res==EXCEPT_LAST && int(head[i]-'0')<=LAST_DIGIT)
				res=res*10+int(head[i]-'0');
			else
				return OVERFLOW_ANSWER;
		}
		return sig==1?res:(-res);
    }
};