LeetCode题目：String to Integer (atoi)

最新推荐文章于 2022-07-27 16:48:14 发布

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最新推荐文章于 2022-07-27 16:48:14 发布

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分类专栏：算法应用作业文章标签： leetcode 算法设计 string

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本文介绍了一种将字符串转换为整数的方法，并提供了一个具体的C++实现案例。该方法考虑了多种输入情况，包括正负号、空格及非法字符等。通过对每个字符的状态进行判断，最终返回对应的整数值。

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题目原址：点击打开链接

题目描述：

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.

我的代码：

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