[LeetCode]atoi

最新推荐文章于 2025-04-28 09:58:53 发布

原创最新推荐文章于 2025-04-28 09:58:53 发布 · 400 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

149 篇文章

订阅专栏

本文介绍如何实现字符串到整数的转换（atoi），详细解释了处理各种输入情况的方法，包括忽略前导空格、判断正负号、解析数字字符等，并确保最终结果符合整数范围限制。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

LeetCode Source

思路：题目不难关键是要注意边界条件。

比如超出Int最大值，最小值。

比如输入的错误。

比如返回是正数还是负数。

空格的处理都要考虑好。

class Solution {
public:
    int atoi(string str) {
        if(str.size()==0)
        return 0;
        long int num = 0;
        int flag = 1;
        auto i = str.begin();
        
        while(*i==' ') i++;
       
        if(*i=='-')  
        {
           flag = -1;
           i++;
        }
        else if(*i=='+')
        {
            flag = 1;
            i++;
        } //'+' or '-'
       
       for(;i!=str.end();++i)
        {
            if(*i>='0'&&*i<='9')
            {
            num=num*10+flag*(*i-'0');
            if(num>=(int)0x7FFFFFFF)
            return 0x7FFFFFFF;
            if(num<=(int)0x80000000)
            return 0x80000000;
            }
            else  break;
        ｝
        return num;
    }
};