C++實現LeetCode(8.字符串轉為整數)

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[LeetCode] 8. String to Integer (atoi) 字符串轉為整數

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.

字符串轉為整數是很常用的一個函數，由於輸入的是字符串，所以需要考慮的情況有很多種。博主之前有一篇文章是關於驗證一個字符串是否為數字的，參見 Valid Number。在那篇文章中，詳細的討論瞭各種情況，包括符號，自然數，小數點的出現位置，判斷他們是否是數字。個人以為這道題也應該有這麼多種情況。但是這題隻需要考慮數字和符號的情況：

1. 若字符串開頭是空格，則跳過所有空格，到第一個非空格字符，如果沒有，則返回0.

2. 若第一個非空格字符是符號 +/-，則標記 sign 的真假，這道題還有個局限性，那就是在 c++ 裡面，+-1 和-+1 都是認可的，都是 -1，而在此題裡，則會返回0.

3. 若下一個字符不是數字，則返回0，完全不考慮小數點和自然數的情況，不過這樣也好，起碼省事瞭不少。

4. 如果下一個字符是數字，則轉為整形存下來，若接下來再有非數字出現，則返回目前的結果。

5. 還需要考慮邊界問題，如果超過瞭整型數的范圍，則用邊界值替代當前值。

C++ 解法：

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

Java 解法：

public class Solution {
    public int myAtoi(String str) {
        if (str.isEmpty()) return 0;
        int sign = 1, base = 0, i = 0, n = str.length();
        while (i < n && str.charAt(i) == ' ') ++i;
        if (i < n && (str.charAt(i) == '+' || str.charAt(i) == '-')) {
            sign = (str.charAt(i++) == '+') ? 1 : -1;
        }
        while (i < n && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
            if (base > Integer.MAX_VALUE / 10 || (base == Integer.MAX_VALUE / 10 && str.charAt(i) - '0' > 7)) {
                return (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            base = 10 * base + (str.charAt(i++) - '0');
        }
        return base * sign;
    }
}

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