[LeetCode]Top Interview Questions/Easy Collection/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.

Solution

#include <iostream>
using namespace std;


class Solution {
public:
    int myAtoi(string str);
};

int Solution::myAtoi(string str){
    int white_space_num = 0; // 开头空格
    // 去除空格
    for(; white_space_num < str.length() ;){
        if(str[white_space_num] == ' '){
            white_space_num ++;
            continue;
        }else{
            break;
        }
    }
    // 全部空格
    if(white_space_num >= str.length()){
        return 0;
    }
    // 数字 或 数字符号
    else if( str[white_space_num] == '-'|| str[white_space_num] == '+' || (str[white_space_num] <= '9' && str[white_space_num] >= '0')){
        bool flag = false; // 负数标记
        int result = 0; // 最终结果
        int index = white_space_num; // 轮询开头
        // 负数符号
        if(str[white_space_num] == '-'){
            flag = true;
            index ++;
        // 正数符号
        }else if(str[white_space_num] == '+'){
            index ++;
        }
        // 去除数字开头0
        while(str[index] == '0' && index <str.length()){
            index ++;
        }
        int start = index;
        int count = 0; // 轮询计数
        while(index < str.length() && str[index] <= '9' && str[index] >= '0' ){
            result = result * 10 + (str[index] - '0');
            index ++;
            count ++;
            // 长度超出10，直接返回
            if(count > 10 ){
                if(flag){
                    return -2147483648;
                }else{
                    return 2147483647;
                }
            }
        }
        if(count == 10 && !flag){ // 位数为10且为正数
            string a = str.substr(start,index);
            string b = "2147483647";
            for(int i = 0 ; i < 10 ;i ++){
                if(a[i]>b[i])
                    return 2147483647;
                else if(a[i]<b[i])
                    return result;
                else
                    continue;
            }
        }else if(count == 10 && flag){// 位数为10且为负数
            string a = str.substr(start,index);
            string b = "2147483648";
            for(int i = 0 ; i < 10 ;i ++){
                if(a[i]>b[i])
                    return -2147483648;
                else if(a[i]<b[i])
                    return -result;
                else
                    continue;
            }
        }
        if(flag){
            result = -result; // 结果为负数
        }
        return result;
    }
    // 开头为非数字符号
    else{
        return 0;
    }
}


int main(int argc, const char * argv[]) {
    Solution solution;
    cout << solution.myAtoi("42") << endl;
    cout << solution.myAtoi("   -42") << endl;
    cout << solution.myAtoi("4193 with words") << endl;
    cout << solution.myAtoi("words and 987") << endl;
    cout << solution.myAtoi("-91283472332") << endl;
    cout << solution.myAtoi("91283472332") << endl;
    cout << solution.myAtoi(" ") << endl;
    cout << solution.myAtoi("-") << endl;
    cout << solution.myAtoi("+1") << endl;
    cout << solution.myAtoi("  0000000000012345678") << endl;
    cout << solution.myAtoi("-000000000000001") << endl;
    cout << solution.myAtoi("-2147483649") << endl;
    cout << solution.myAtoi("2147483648") << endl;
    cout << solution.myAtoi("-2147483647") << endl;
    cout << solution.myAtoi("-6147483648") << endl;
    return 0;
}