leetcode 8 string to integer(atoi)

最新推荐文章于 2021-02-26 22:34:04 发布

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#leetcode 8 string to

本文详细解析 LeetCode 上字符串转整数 (atoi) 的题目要求，并提供了一种清晰的 C++ 实现方案，包括去除空白字符、处理正负号及数值范围限制。

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题目链接：https://leetcode.com/problems/string-to-integer-atoi/

题目的意思很是简单：Implement atoi to convert a string to an integer，实现函数 atoi。

如果对atoi函数不是很了解的话，还真是不怎么好写。不过还好有比较多的提示。参照提示来写就是比较有思路了。

提示的话，放这儿翻译一下：

/*
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.

首先得去掉字符串的开头空格。第一个可能为+或者-，需要注意。

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.

在数字的最后可能有一些无用的字符，我们应该无视它们。

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.

如果第一个字符是非法的，或者如果str的空的话，这个字符串的非法的。

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.

对于非法的字符串应该返回0，如果有超过数据范围的话，返回边界值。
*/

Code：

class Solution {
public:
    int myAtoi(string str) {
        int len = str.size();
        int l = 0, r = len - 1;
        /*
         * find the left and the right of the int.
         * than count the answer.
        */
        while(str[l] == ' ') ++ l;
        if(str[l] == '+' || str[l] == '-') l ++;

        while(!(str[r] >= '0' && str[r] <= '9')) -- r;
        for(int i = l; i <= r; ++ i){
            if(!(str[i] >= '0' && str[i] <= '9')) r = i - 1;
        }
        /*
         * special for the case:
        */
        if(l > r) return 0;
        if(r - l + 1 >= 15){
            if(str[l - 1] == '-') {
                return  -2147483648;
            }
            else return 2147483647;
        }

        long long ans = 0, k = 1;
        for(int i = r; i >= l; -- i){
            ans += (str[i] - '0') * k; 
            k = k * 10;
        }

        if(l != 0 && str[l - 1] == '-') ans = -ans;
        if(ans > 2147483647) ans = 2147483647;
        if(ans < -2147483648) ans = -2147483648;

        return (int)ans;
    }
};