Ray's work

[leetcode]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.

容易忽略的case， +123, overflow

public class atoi {

    public int charToNum(char c){
        int num = c - '0';
        if(num >= 0 && num <= 9)return num;
        return 0;
    }

    public int atoi(String s) {
        s = s.trim();
        long ans = 0L;
        boolean isNegative = false;
        if (s.length() == 0) return 0;
        if (s.length() == 1) return charToNum(s.charAt(0));
        if (s.charAt(0) == '-') isNegative = true;
        int pos = 0;
        if (isNegative||s.charAt(0) == '+')pos++;

        while ((pos < s.length()) && s.charAt(pos) >= '0' && s.charAt(pos) <= '9') {
            ans = ans * 10 + (s.charAt(pos) - '0');
            pos++;
            //Negative
            if (ans > 0x7fffffff) return isNegative ? Integer.MIN_VALUE : Integer.MAX_VALUE;//overflow check
        }
        return isNegative ? -(int) ans : (int) ans;
    }

    public static void main(String[] args) {
        atoi a = new atoi();
        String[] arr = {"   -123", "12 3", "abcd", "-", "12321234123412431243124","   010"};
        for (int i = 0; i < arr.length; i++) {
            System.out.println(a.atoi(arr[i]));
        }
    }
}

leetcode solution还是比较clean的，有几点值得借鉴

  1. overflow check可以用MAX/10来比较，省的定义输入为long，还要强转
  2. 正负可以用一个sign，省去了return用表达式

Code:

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;
}
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Published

25 November 2014

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