String to Integer (week2)

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.
<strong>Note</strong>:

• 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: [, ]. 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

Example2:

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.

Example3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example4:

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.

Example5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.Thefore INT_MIN ($−2^{31}$) is returned.

解题分析

这是一道水题，直接遍历整个字符串根据条件判断即可。

1. 当还未出现数字或正负号时，允许空格和正负号
2. 当出现了数字之后，将数字放入result中。
3. 当出现了数字之后，出现非数字的内容，则跳出，并返回result
4. 当出现的第一个字符是字母或者其他非数字、空格、正负号字符时，直接返回0

时间复杂度分析

显然，遍历一遍字符串是O(n)的复杂度。

源码

class Solution {
public:
    int myAtoi(string str) {
       int _INT_MAX = 2147483647, _INT_MIN = -2147483648;
        long long result = 0;
        bool opposite = false, begin = false;
        for(int i = 0; i < str.length(); ++ i) {
            if (result > _INT_MAX || result < _INT_MIN)
                break;
            if (!begin && str[i] == ' ') {
                continue;
            } else if (!begin && str[i] == '-') {
                opposite = true;
                begin = true;
            } else if (!begin && str[i] == '+') {
                begin = true;
            } else if (str[i] >= '0' && str[i] <= '9') {
                begin = true;
                result = result * 10 + str[i] - '0';
            } else if (begin && (str[i] < '0' || str[i] > '9')) {
                break;
            } else {
                result = 0;
                break;
            }
        }
        if (opposite) result = -result;
        if (result >= _INT_MIN && result <= _INT_MAX)
            return result;
        else if (result < _INT_MIN)
            return _INT_MIN;
        else return _INT_MAX;
    }
};