String to Integer

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.

spoilers alert... click to show requirements for atoi.

这道题我觉得挺无趣的，可能是因为OJ题目做的太少，不太清楚各种规则，有些想当然，没有完全理解清楚题意就开始写代码以至于浪费了很长时间。不过有意思的是，当我提交成功之后，系统提示我解锁了一个solution，看了一下官方给的solution，瞬间明白了，好的算法和烂算法的差别。Although I make it possible， I make it ugly。

以下是我冗长的Python代码：

ef atoi(strg):    illeFigure = False    fFigure = False    figure = []    numList = []    num = []    for i  in range(0, 10):        numList.append(str(i))    for i in range(0, len(strg)):        if illeFigure == False:            if strg[i] == " ":                continue            elif strg[i] == "+" or strg[i] == "-":                illeFigure = True                fFigure = True                figure.append(strg[i])                continue            elif strg[i] in numList:                num.append(strg[i])                illeFigure = True                continue            else:                break        if illeFigure == True:               if strg[i] in numList:                num.append(strg[i])                continue            else:                break    if illeFigure == False:        return 0    else:        if len(num) == 0:            return 0        d = ''.join(num)        y = re.findall("[1-9][0-9]*", d)        if len(y) == 0:            return 0        x = ''.join(y)        print("x is:" + x)        if fFigure == True:            if figure[0] == "+":                if len(x) == 10 and x >= "2147483647":                    return 2147483647                elif len(x) > 10:                    return 2147483647                else:                    return int(x)            if figure[0] == "-":                if len(x) == 10 and x >= "2147483648":                    return -2147483648                elif len(x) > 10:                    return -2147483648                else:                    return int(-int(x))        if len(x) == 10 and x > "2147483647":            return 2147483647        elif len(x) > 10:            return 2147483647        else:            return int(x)

官方高大上的solution：

To deal with overflow, inspect the current number before multiplication. If the current number is greater than 214748364, we know it is going to overflow. On the other hand, if the current number is equal to 214748364, we know that it will overflow only when the current digit is greater than or equal to 8.private static final int maxDiv10 = Integer.MAX_VALUE / 10;public int atoi(String str) {   int n = str.length();   int i = 0;   while (i < n && Character.isWhitespace(str.charAt(i))) i++;   int sign = 1;   if (i < n && str.charAt(i) == '+') {      i++;   } else if (i < n && str.charAt(i) == '-') {      sign = -1;      i++;   }   int num = 0;   while (i < n && Character.isDigit(str.charAt(i))) {      int digit = Character.getNumericValue(str.charAt(i));      if (num > maxDiv10 || num == maxDiv10 && digit >= 8) {         return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;      }      num = num * 10 + digit;      i++;   }     return sign * num;}