[LeetCode] 008. String to Integer (Easy) (C++/Java/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)

Github: https://github.com/illuz/leetcode

008.String_to_Integer (Easy)

链接：

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

代码(github)：https://github.com/illuz/leetcode

题意：

将一个字符串转化为 int 型。

分析：

注意如果超出范围就返回最接近的 int 数。

如：2147483648 大于 INT_MAX(2147483647) ，就返回 2147483647 。

之前可以用 sscanf 偷懒，最近更新了 case 就被卡了。

要注意几点：

1. 跳过前面的空格,\t,\n
2. 范围界定

使用 Python 的正则表达式可以很容易处理。

代码：

C++:

class Solution {
public:
    int atoi(string str) {
		int ret = 0;
		bool overflow = false;
		int sign = 1;	// default is '+'
		int i = 0;
		int len = str.length();

		while (i < len && (str[i] == ' ' || str[i] == '\n' || str[i] == '\t'))
			++i;
		if (i == len)
			return 0;

		// get sign
		if (str[i] == '-') {
			++i;
			sign = -1;
		} else if (str[i] == '+')
			++i;

		while (i < len) {
			if (!isdigit(str[i]))
				break;
			if ((sign == 1 && ret > (INT_MAX - (str[i]-'0')) / 10) ||
					(sign == -1 && -ret < (INT_MIN + (str[i]-'0')) / 10)) {
				overflow = true;
				break;
			}
			ret = ret * 10 + (str[i] - '0');
			++i;
		}
		if (overflow)
			ret = (sign == 1) ?  INT_MAX : INT_MIN;
		else
			ret *= sign;
		return ret;
    }
};

Java:

public class Solution {

    public int atoi(String str) {

        int ret = 0;
        boolean overflow = false;
        int sign = 1;	// default is '+'
        int i = 0;
        int len = str.length();

        while (i < len && (str.charAt(i) == ' ' || str.charAt(i) == '\n' || str.charAt(i) == '\t'))
            ++i;
        if (i == len)
            return 0;

        // get sign
        if (str.charAt(i) == '-') {
            ++i;
            sign = -1;
        } else if (str.charAt(i) == '+')
            ++i;

        while (i < len) {
            if (str.charAt(i) < '0' || str.charAt(i) > '9')
                break;
            if ((sign == 1 && ret > (Integer.MAX_VALUE - (str.charAt(i)-'0')) / 10) ||
                    (sign == -1 && -ret < (Integer.MIN_VALUE + (str.charAt(i)-'0')) / 10)) {
                overflow = true;
                break;
            }
            ret = ret * 10 + (str.charAt(i) - '0');
            ++i;
        }
        if (overflow)
            ret = (sign == 1) ?  Integer.MAX_VALUE: Integer.MIN_VALUE;
        else
            ret *= sign;
        return ret;
    }
}

Python:

class Solution:
    # @return an integer
    def atoi(self, str):
        str = str.strip()
        if not str:
            return 0

        MAX_INT = 2147483647
        MIN_INT = -2147483648
        ret = 0
        overflow = False
        pos = 0
        sign = 1

        if str[pos] == '-':
            pos += 1
            sign = -1
        elif str[pos] == '+':
            pos += 1

        for i in range(pos, len(str)):
            if not str[i].isdigit():
                break
            ret = ret * 10 + int(str[i])
            if not MIN_INT <= sign * ret <= MAX_INT:
                overflow = True
                break

        if overflow:
            return MAX_INT if sign == 1 else MIN_INT
        else:
            return sign * ret

使用 Python 的正则表达式:

class Solution:
    # @return an integer
    def atoi(self, str):
        str = str.strip()
        str = re.match(r'^[+-]?\d+', str).group()
        MAX_INT = 2147483647
        MIN_INT = -2147483648

        try:
            ret = int(str)
            if ret > MAX_INT:
                return MAX_INT
            elif ret < MIN_INT:
                return MIN_INT
            else:
                return ret
        except:
            return 0