8. String to Integer (atoi)

最新推荐文章于 2020-04-03 12:34:44 发布

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本文介绍了一种将字符串转换为整数的有效方法，并通过两个不同的C++实现展示了如何正确处理边界条件，避免整数溢出的问题。

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.

class Solution {
public:
    int myAtoi(string str) {
    int sign = 1, i = 0, ans = 0;
    while(str[i] == ' ') ++i;
    if(str[i] == '-' || str[i] == '+') 
    {   
        sign = 1 - 2*(str[i++] == '-');
    }
    while(str[i] >= '0' && str[i] <= '9')
    {
        if (ans >  INT_MAX / 10 || (ans == INT_MAX / 10 && str[i] - '0' > 7)) {
            if (sign == 1) return INT_MAX;
            else return INT_MIN;
        } 
        ans  = 10 * ans + (str[i++] - '0');  
        
    }
    return  ans * sign;
    
    
    
    }
};

这两个方法一个先判断是否越界，一个后判断是否越界，下面的后判断是否越界的，如果ans定义成int 然后后判断，会出现错误！先溢出！所以下面定义成long long 才可以！

class Solution {
public:
    int myAtoi(string str) {
    int sign = 1, i = 0;
    long long ans = 0;
    while(str[i] == ' ') ++i;
    if(str[i] == '-' || str[i] == '+') 
    {   
        sign = 1 - 2*(str[i++] == '-');
    }
    while(str[i] >= '0' && str[i] <= '9')
    {
         
        ans  = 10 * ans + (str[i++] - '0');  
        if (ans >  INT_MAX ) {
            if (sign == 1) return INT_MAX;
            else return INT_MIN;
        }
        
    }
    return  ans * sign;
    
    
    
    }
};