【Leetcode】Additive Number

题目链接：https://leetcode.com/problems/additive-number/ 题目：

Additive number is a string whose digits can form additive sequence.

at least

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, 3'0'-'9', write a function to determine if it's an additive number.

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

思路：

如果是additive number，则n1+n2=n3，n2+n3一定在字符串中，即可以递推，若不在，则不是additive number。

所以思路就是遍历找到第一、二个数，看它们是否满足上述条件。前导零的问题，以前做题的时候也遇到过，注意即可。

BigInteger要自己手动再AC代码前import java.math.BigInteger 否则会找不到此类。

算法：

public boolean isAdditiveNumber(String num) {
		for(int i=1;i<num.length();i++){
			if(i>1&&num.charAt(0)=='0')
				return false;
			BigInteger n1 = new BigInteger(num.substring(0,i));
			for(int j=i+1;j<num.length();j++){
				BigInteger n2 = new BigInteger(num.substring(i,j));
				if(j>i+1&&num.charAt(i)=='0')
					break;
				if(isvalid(n1,n2,num.substring(j))){
					return true;
				}
			}
		}
		return false;
	}
	

	private boolean isvalid(BigInteger n1, BigInteger n2, String num) {
		if(num.length()==0)
			return true;
		BigInteger n3 = n2.add(n1);
		return num.startsWith(n3.toString())&&isvalid(n2,n3,num.substring(n3.toString().length()));
	}