8. String to Integer (atoi)-优快云博客

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/79861389

本文介绍如何实现字符串转整数的atoi方法，包括处理前导空格、符号位、溢出及非法输入等问题，提供了一个简洁的C++实现。

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问题描述

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.

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.

题目链接：

思路分析

实现string转int的atoi()方法。

一道很操蛋的题目，我们需要考虑四种情况：1.字符前面有空格；2.符号位；3.溢出；4.非法输入（++-93）。

依次解决，跳过空格；判断符号位；设置循环条件（string最后一位是‘/n’，其他的非法输入也不会触发循环；判断溢出，是否大于INT_MAX的十分之一或者正好大于一丢丢。

代码

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