LeetCode 8. String to Integer (atoi)

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).

The algorithm for myAtoi(string s) is as follows:

Read in and ignore any leading whitespace.
Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
Return the integer as the final result.

Note:

Only the space character ' ' is considered a whitespace character.
Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

Example 1:

Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
		 ^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
		 ^
Step 3: "42" ("42" is read in)
		   ^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.

Example 2:

Input: s = "   -42"
Output: -42
Explanation:
Step 1: "   -42" (leading whitespace is read and ignored)
			^
Step 2: "   -42" ('-' is read, so the result should be negative)
			 ^
Step 3: "   -42" ("42" is read in)
			   ^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.

Example 3:

Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
		 ^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
		 ^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
			 ^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.

Example 4:

Input: s = "words and 987"
Output: 0
Explanation:
Step 1: "words and 987" (no characters read because there is no leading whitespace)
		 ^
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
		 ^
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
		 ^
The parsed integer is 0 because no digits were read.
Since 0 is in the range [-231, 231 - 1], the final result is 0.

Example 5:

Input: s = "-91283472332"
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
		 ^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
		  ^
Step 3: "-91283472332" ("91283472332" is read in)
					 ^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.

Constraints:

0 <= s.length <= 200
s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.

实现思路：

本题就算要求实现一个库函数atoi()的功能，leetcode官方采用了有限状态自动机的方法，emmmm自己比较菜，先实现基本方法吧，至于其他方法，留到以后再次优化自己代码的时候再进行扩充。

AC代码：

class Solution {
	public:
		int myAtoi(string s) {
			while(*s.begin()==' ') s.erase(s.begin());//消除前缀空格
			int MAX=0x7fffffff,MIN=0x80000000,res=0;//定义最大和最小值
			string ans;
			if(s[0]=='-') {//有符号的话进行删去
				ans+='-';
				s.erase(s.begin());
			} else if(s[0]=='+') s.erase(s.begin());
			for(int i=0; i<s.length(); i++) {
				if(!(s[i]>='0'&&s[i]<='9')) break;//遇到非数字字母就退出循环
				ans+=s[i];
			}
			long long val=atoll(ans.c_str());//这里数据会超过int 所以用long long来接收
			if(val>MAX) res=MAX;//如果大于int的数据范围了就进行数据截断
			else if(val<MIN)
				res=MIN;
			else res=val;
			return res;
		}
};

tips：本处最大最小值用的是MAX=0x7fffffff,MIN=0x80000000，本来最大值我定义成MAX=(1<<31)-1，也就是用位移方式去定义，但是还是出错了，这就不知道为什么了，如果有知道的小伙伴可以讨论一下。