字符串转整数atoi实现-优快云博客

原题链接：https://leetcode.com/problems/string-to-integer-atoi/#/description

题目：

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.

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.

题目说明：

实现一个函数，将字符串转换成数字，例如将"123"转换成123。由于输入字符串形式并未指定，所以需要考虑以下五种情况：

1.从第一个非空字符串开始处理。如s=" __123" (_代表空格键）则应该从第三个字符开始处理。

2.要处理字符串前面的“”“”

3.字符串中可能含有非'0'~'9'的字符，遇到这样的字符，该字符后面的字符失效，例如：s="12a3" 应转换为 12.

4.若该字符串中全是空格，则函数不进行任何转换。

5.越界情况，当转换后数字超高int 的最大值或最小值时，返回int型的最大值或最小值 INT_MAX (2147483647) or INT_MIN (-2147483648)

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

处理空格比较简单，第一个for循环可以找到第一个非空格的字符。由于涉及到'+'和'-'，所以设置sign符号位，用来表示第一个非空格字符是'+'还是'-'.

在wihle循环的执行条件是字符必须在'0'~'9'之间，且下标没有超出字符串长度，在循环体中，将字符转换成数字。并通过语句

if(sign*result>=INT_MAX) return INT_MAX;
if(sign*result<=INT_MIN) return INT_MIN;

判断是否越界。注意此处的result为long long 类型，其可以表示比INT_MAX大或比INT_MIN小的数字。

另外，看到一段很巧妙的判断是否越界的代码：

int atoi(const char *str) {
    int sign = 1, base = 0, i = 0;
    while (str[i] == ' ') { i++; }
    if (str[i] == '-' || str[i] == '+') {
        sign = 1 - 2 * (str[i++] == '-'); 
    }
    while (str[i] >= '0' && str[i] <= '9') {
        if (base >  INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7)) {
            if (sign == 1) return INT_MAX;
            else return INT_MIN;
        }
        base  = 10 * base + (str[i++] - '0');
    }
    return base * sign;
}

该段程序通过语句

base >  INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - '0' > 7

判断base是否越界，经典的是str[i]-'0'>7；这是因为INT_MAX最大为 2147483647，INT_MAX/10=214748364,最后的个位数字是7.所以只要str[i]-'0'>7那么base肯定就是越界了。。。。

补充知识点：

1.关于INT_MAX与INT_MIN(https://www.quora.com/What-is-INT_MIN-and-INT_MAX-in-C++)

INT_MIN is a macro that expands to the smallest (most negative) value that can be stored in a variable of type int.

INT_MAX is a macro that expands to the largest (most positive) value that can be stored in an int.

On most processors INT_MIN == -INT_MAX - 1, i.e., there is one more negative number than positive number in the range of legal values.

For unsigned int, the corresponding values are 0 and UINT_MAX. Typically UINT_MAX == 2*INT_MAX+1.

Leetcode--8

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