Additive Number
<p>An <strong>additive number</strong> is a string whose digits can form an <strong>additive sequence</strong>.</p>

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

<p>Given a string containing only digits, return <code>true</code> if it is an <strong>additive number</strong> or <code>false</code> otherwise.</p>

<p><strong>Note:</strong> Numbers in the additive sequence <strong>cannot</strong> have leading zeros, so sequence <code>1, 2, 03</code> or <code>1, 02, 3</code> is invalid.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> &quot;112358&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> 
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
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> &quot;199100199&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> 
The additive sequence is: 1, 99, 100, 199.&nbsp;
1 + 99 = 100, 99 + 100 = 199
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num.length &lt;= 35</code></li>
	<li><code>num</code> consists only of digits.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> How would you handle overflow for very large input integers?</p>
String; Backtracking
class Solution {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        for (int i = 1; i < Math.min(n - 1, 19); ++i) {
            for (int j = i + 1; j < Math.min(n, i + 19); ++j) {
                if (i > 1 && num.charAt(0) == '0') {
                    break;
                }
                if (j - i > 1 && num.charAt(i) == '0') {
                    continue;
                }
                long a = Long.parseLong(num.substring(0, i));
                long b = Long.parseLong(num.substring(i, j));
                if (dfs(a, b, num.substring(j))) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean dfs(long a, long b, String num) {
        if ("".equals(num)) {
            return true;
        }
        if (a + b > 0 && num.charAt(0) == '0') {
            return false;
        }
        for (int i = 1; i < Math.min(num.length() + 1, 19); ++i) {
            if (a + b == Long.parseLong(num.substring(0, i))) {
                if (dfs(b, a + b, num.substring(i))) {
                    return true;
                }
            }
        }
        return false;
    }
}