leetcode 8 -- String to Integer (atoi)

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本文详细介绍了如何实现字符串到整数的转换算法，包括处理字符串开头的空格、正负号以及防止数值越界等问题。通过实例代码，深入探讨了atoi函数的工作原理及注意事项。

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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.

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.

题意

将字符串转化为整形数字输出
注意三个问题：
一是字符串开头空格的问题
二是正负号
数值越界问题（题意已提到若越界则输出INT-MAX 或 INT-MIN）

思路

此题较简单，不做赘述，见代码注释

代码

int myAtoi(char* str) {
    int i=0;
    long long rel;//考虑到越界问题用longlong保存
    while(str[i]==' '){i++;}
    //消除空格
    rel=atol(str);//此处取巧了 用了atol
    if(rel>INT_MAX)//处理越界
    {
        return INT_MAX;
    }
    else if(rel<INT_MIN)
    {
        return INT_MIN;
    }
    return rel;
}