LeetCode Additive Number

Description:

Additive number is a string 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 containing only digits '0'-'9', write a function to determine if it's an additive number.

Solution:

还是一道不给数据范围就尝试DFS的题目……

我这里用的方法比较暴力，还有一种更简单的，因为每次符合additive number的方法只要确定前两个数字就好

<span style="font-size:18px;">&& i < tot + 10</span>

中加有一步加了这个判断，意思就是超过10位的数字我们就不考虑了，实际上是个bug，万一BigInteger呢？

但再考虑到题目说得这么不严谨……

<span style="font-size:18px;">public class Solution {
	public boolean isAdditiveNumber(String num) {
		int n = num.length();
		long[] arr = new long[n];
		return dfs(0, 0, num, arr);

	}

	public boolean dfs(int tot, int current, String num, long[] arr) {
		if (tot == num.length()) {
			for (int i = 2; i < current; i++)
				if (arr[i - 2] + arr[i - 1] != arr[i])
					return false;
			if (current <= 2)
				return false;
			return true;
		}

		if (current >= 3
				&& arr[current - 3] + arr[current - 2] != arr[current - 1])
			return false;

		for (int i = tot; i < num.length() && i < tot + 10; i++) {
			long current_num = Long.parseLong(num.substring(tot, i + 1));
			arr[current] = current_num;
			if (num.charAt(tot) == '0' && current_num > 0)
				return false;
			if (dfs(i + 1, current + 1, num, arr))
				return true;
		}

		return false;
	}

}</span>