LeetCode-String to Integer (atoi)

String to Integer (atoi)

##题目

####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.

####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.

##解题思路
该题是将字符串转换为整数，但是其中结束条件比较多。需要考虑正负符号问题和整数越界问题。

####该算法的求解步骤如下：

1. 先去掉多余的空格字符，从第一个非空格字符开始。
2. 按顺序读数字，如果出现下面三种情况则结束：
    2.1 异常字符出现（按照C语言的标准是把异常字符后面的全部截掉，保留前面的部分作为结果）。
    2.2 数字越界（返回最接近的整数）。
    2.3 字符串结束

##算法代码
代码采用JAVA实现：

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public class Solution {
    public int atoi(String str) {
        char[] c=str.toCharArray();
        int len=c.length;
        if(len==0) return 0;
        int i=0;
        int sign=1;
        int num=0;
        while(c[i]==' '&&i<len) //从第一个非空格字符开始
        	i++;
        if(i>=len) return 0;
        if(c[i]=='+')   //判断整数的符号，如果没有正负号，则默认为正
        {
        	sign=1;
        	i++;
        }else if(c[i]=='-')
            {
            	sign=-1;
            	i++;
            }
        for(;i<len;i++)
        {
        	if(c[i]<'0'||c[i]>'9') //如果出现异常字符则退出
        		break;
        	if((num>Integer.MAX_VALUE/10)||((num==Integer.MAX_VALUE/10)&&((c[i]-'0')>Integer.MAX_VALUE%10))) //如果出现数值越界，则返回最接近的整数
        	{
        		if(sign==1)
        			return Integer.MAX_VALUE;
        		else
        			return Integer.MIN_VALUE;
        	}
        	num=num*10+(c[i]-'0');
        }
        return sign*num;
    }
}

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