[LeetCode] (medium) 8. String to Integer (atoi)-优快云博客

本文链接：https://blog.youkuaiyun.com/Frontier_Setter/article/details/86746979

本文详细解析了LeetCode上字符串转整数(Atoi)题目的实现过程，重点介绍了如何处理字符串中的空白字符、正负号以及数字转换，并讨论了溢出检测的方法。文章通过具体示例解释了算法的工作原理，提供了C++代码实现。

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

Implement atoi which converts a string to an integer.

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.

Note:

Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.

Example 1:
Input: "42"
Output: 42
Example 2:
Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.
Example 3:
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Example 4:
Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.
Example 5:
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−231) is returned.

难点在于溢出判断，参考了以下方法：

https://blog.youkuaiyun.com/pizi0475/article/details/45721171

对于有符号整数的溢出，只需要简单判断运算结果符号是否与操作数相等即可；下面我们讨论无符号integer的溢出检测问题：

假设我们有两个变量a和b，size 为n，最大值为R。'+'代表实际的数学运算符--加号，‘$’代表计算机中的运算。

加法溢出

显然如果a+b<=R-1;a$b=a+b;

如果a+b>=R;a$b=a+b-R; 分析：由于R比a和b都大，所以a-R与b-R都是负值，所以我们有a+b-R<a;a+b-R<b;

乘法溢出

1.如果a*b>max，则有a>max/b;(如果无符号，max=R-1;否则max=R/2-1)

注意在加法溢出判断的时候不能用

if (result+(str[cur]-'0') >= result){}

可能因为编译器的优化，达不到检测的效果

class Solution {
public:
    int myAtoi(string str) {
        static int fast_io = []() { std::ios::sync_with_stdio(false); cin.tie(nullptr); return 0; }();
        int cur = 0;
        int result = 0;
        int len = str.size();
        bool neg = false;
        int thre = INT_MAX/10.0;
        //cout << thre << endl;
        while(cur < len && str[cur] == ' ') ++cur;
        
        if(cur == len) return result;
        
        if(str[cur] == '-'){
            neg = true;
            ++cur;
        }else if(str[cur] == '+'){
            ++cur;
        }
        if(cur == len) return result;
        
        while(cur < len && (str[cur] >= '0' && str[cur] <= '9')){
            //cout << "detec: " << str[cur] << " " << result << endl;
            if((float)result <= thre){
                result *= 10;
            }else{
                if(!neg) return INT_MAX;
                else return INT_MIN;
            }
            result = result+(str[cur]-'0');
            if(result < 0){
                if(!neg) return INT_MAX;
                else return INT_MIN;
            }
            ++cur;
        }
        if(neg) result = -result;
        return result;
    }
};