LeetCode 题解(290): Additive Number

题目：

Additive number is a positive integer whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

For example:
"112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8.

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

"199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.

1 + 99 = 100, 99 + 100 = 199

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Given a string represents an integer, write a function to determine if it's an additive number.

Follow up:
How would you handle overflow for very large input integers?

题解：

Java版：

public class Solution {
    public boolean isAdditiveNumber(String num) {
        //枚举前两个数的位置，因为前两个数决定了num是否为Additive
        for(int i = 0; i < num.length(); i++) {
            for(int j = i + 1; j < num.length() - i - 1; j++) {
                String first = num.substring(0, i + 1);
                String second = num.substring(i + 1, j + 1);
                if(valid(j + 1, num, first, second))
                    return true;
            }
        }
        return false;
    }

    private boolean valid(int start, String num, String first, String second) {
        if(start == num.length())
            return true;
        long f = Long.parseLong(first);
        long s = Long.parseLong(second);
        if(!Long.toString(f).equals(first) || !Long.toString(s).equals(second))
            return false;
        long sum = f + s;
        String sumS = Long.toString(sum);
        if(start + sumS.length() > num.length())
            return false;
        String third = num.substring(start, start + sumS.length());
        long t = Long.parseLong(third);
        if(!Long.toString(t).equals(third) || t != sum)
            return false;
        return valid(start + sumS.length(), num, second, third);
    }
}