蜗牛慢慢爬 LeetCode 8. String to Integer (atoi) [Difficulty: Medium]

题目

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Requirements for atoi:

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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

翻译

实现atoi将字符串转换为整数。

提示：仔细考虑所有可能的输入案例。如果你想要一个挑战，请不要在下面看到，问自己什么是可能的输入案例。

注意：这个问题的目的是模糊地指定（即没有给定的输入规范）。您有责任收集所有的输入要求。

atoi的要求：

该函数首先丢弃尽可能多的空格字符，直到找到第一个非空格字符。然后，从该字符开始，选择一个可选的初始加号或减号，后跟尽可能多的数字，并将其解释为数值。

该字符串可以包含形成整数之后的其他字符，这些字符将被忽略，并且不影响此函数的行为。

如果str中的第一个非空白字符序列不是有效的整数，或者由于str为空或仅包含空格字符，否则不存在此类序列，则不进行转换。

如果不能执行有效的转换，则返回零值。如果正确的值超出可表示值的范围，则返回INT_MAX（2147483647）或INT_MIN（-2147483648）。

Hints

Related Topics: Math, String

代码

Java

class Solution {
    public int myAtoi(String str) {
        int sign = 1;
        int base = 0;
        int i = 0;
        int INT_MAX = 2147483647;
        int INT_MIN = -2147483648;
        if(str.length()==0) return 0;
        while(str.charAt(i)==' ')   i++;
        if(str.charAt(i)=='+'||str.charAt(i)=='-'){
            sign = str.charAt(i++)=='+'?1:-1;
        }
        while(i<str.length()&&str.charAt(i)<='9'&&str.charAt(i)>='0'){
            if(base > INT_MAX/10||(base==INT_MAX/10 && (str.charAt(i)-'0')>7)){
                if(sign==1) return INT_MAX;
                else    return INT_MIN;
            }
            base = 10*base + (str.charAt(i++)-'0');
        }
        return base*sign;
    }
}

Python

class Solution(object):
    def myAtoi(self, str):
        """
        :type str: str
        :rtype: int
        """
        i = base = 0
        sign = 1
        INT_MAX = 2147483647
        INT_MIN = -2147483648
        if len(str)==0:
            return 0
        while str[i]==' ':
            i += 1
        if str[i]=='-' or str[i]=='+':
            sign = 1 if str[i]=='+' else -1
            i += 1
        while i<len(str) and str[i]<='9' and str[i]>='0':
            if base > INT_MAX/10 or (base==INT_MAX/10 and int(str[i])>7):
                if sign==1:return INT_MAX
                else: return INT_MIN
            base = 10*base + int(str[i])
            i += 1
        return base*sign

class Solution:
    def myAtoi(self, str):
        str = str.strip()
        if len(str) == 0:
            return 0
        tmp = "0"
        result = 0
        i = 0
        if str[0] in "+-":
            tmp = str[0]
            i = 1
        MAX_INT = 2147483647
        MIN_INT = -2147483648
        for i in xrange(i, len(str)):
            if str[i].isdigit():
                tmp += str[i]
            else:
                break
        if len(tmp) > 1:
            result = int(tmp)
        if result > MAX_INT > 0:
            return MAX_INT
        elif result < MIN_INT < 0:
            return MIN_INT
        else:
            return result