LeetCode-StringToInteger

题目描述

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.

思路

该题需要去掉字符串前导的空格，去除空格后判断正负号，注意：去完正负号后需要去掉前导的数字0，由于题目的限制不可输出无限大的数，那么定义一个比int类型大的数的位数为maxDigits，然后遍历数字转换为值直到出现非法字符或者超过最大位数，最后只需判断结果是否比int大，按题意输出即可。
这里的将数字串转为值的递推式：

程序代码

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int myAtoi(string str){
    int i = 0;
    while(str[i] == ' ') i++;
    bool negative = false;
    if(str[i] == '+'){
        i++;

    }else if(str[i] == '-'){
        negative = true;
        i++;
    }else if(str[i] <'0' && str[i] > '9'){
        printf("0\n");
        return 0;
    }
    while(str[i] == '0') i++;
    int maxDigits = 12;
    int currDigits = 0;
    long long res = 0;
    while(str[i] >= '0' && str[i] <='9' && currDigits < maxDigits){
        res = res * 10 + (str[i] - '0');
        currDigits ++;
        i++;
    }
    if(negative){
        res = -res;
        if(res < INT_MIN)
            return INT_MIN;
        else return res;
    }else{
        if(res > INT_MAX)
            return INT_MAX;
        else return res;
    }
}

int main(int argc, char const *argv[])
{
    string str;
    cin>>str;
    printf("%d\n", myAtoi(str));
    return 0;
}