<p>You are given a string of digits <code>num</code>, such as <code>&quot;123456579&quot;</code>. We can split it into a Fibonacci-like sequence <code>[123, 456, 579]</code>.</p>

<p>Formally, a <strong>Fibonacci-like</strong> sequence is a list <code>f</code> of non-negative integers such that:</p>

<ul>
	<li><code>0 &lt;= f[i] &lt; 2<sup>31</sup></code>, (that is, each integer fits in a <strong>32-bit</strong> signed integer type),</li>
	<li><code>f.length &gt;= 3</code>, and</li>
	<li><code>f[i] + f[i + 1] == f[i + 2]</code> for all <code>0 &lt;= i &lt; f.length - 2</code>.</li>
</ul>

<p>Note that when splitting the string into pieces, each piece must not have extra leading zeroes, except if the piece is the number <code>0</code> itself.</p>

<p>Return any Fibonacci-like sequence split from <code>num</code>, or return <code>[]</code> if it cannot be done.</p>

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

<pre>
<strong>Input:</strong> num = &quot;1101111&quot;
<strong>Output:</strong> [11,0,11,11]
<strong>Explanation:</strong> The output [110, 1, 111] would also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> num = &quot;112358130&quot;
<strong>Output:</strong> []
<strong>Explanation:</strong> The task is impossible.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;0123&quot;
<strong>Output:</strong> []
<strong>Explanation:</strong> Leading zeroes are not allowed, so &quot;01&quot;, &quot;2&quot;, &quot;3&quot; is not valid.
</pre>

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

<ul>
	<li><code>1 &lt;= num.length &lt;= 200</code></li>
	<li><code>num</code> contains only digits.</li>
</ul>

class Solution {
public:
    vector<int> splitIntoFibonacci(string num) {
        if(num.size()<3)
            return vector<int>();
        vector<int> ans;
        vector<int> tmp;
        bool found = false;
        function<void(int)> backtracking = [&](int idx)
        {
            if(found)
                return;
            if (idx == num.size() && tmp.size() >= 3)
            {
                ans = tmp;
                found = true;
                return;
            }
            long temp = 0;
            for(int i=1; i<=num.size()-idx; ++i)
            {
                if(num[idx] == '0' && i > 1)
                    break;
                temp = 10 * temp + num[idx+i-1] - '0';
                if(temp > INT_MAX)
                    break;
                if(tmp.size() < 2)
                {
                    tmp.push_back(temp);
                    backtracking(idx+i);
                    tmp.pop_back();
                }
                else
                {
                    long lag1 = tmp[tmp.size()-1];
                    long lag2 = tmp[tmp.size()-2];
                    long target = lag1 + lag2;
                    if(temp == target )
                    {
                        tmp.push_back(temp);
                        backtracking(idx+i);
                        tmp.pop_back();
                    }
                    else if (temp > target)
                        break;
                }
            }
        };
        backtracking(0);
        return ans;
    }
};
8400