<p>Given a positive integer <code>n</code>, return the number of the integers in the range <code>[0, n]</code> whose binary representations <strong>do not</strong> contain consecutive ones.</p>

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

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 5
<strong>Explanation:</strong>
Here are the non-negative integers &lt;= 5 with their corresponding binary representations:
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. 
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int findIntegers(int n) {
        int num_digits = 0;
        vector<int> digits;
        while(n != 0){
            digits.push_back(n & 1);
            n = n >> 1;
            num_digits++;
        }
        vector<int> dp;
        dp.push_back(digits[0]+1);
        vector<int> fib;
        fib.push_back(1); fib.push_back(2);
        for(int i = 1; i < num_digits; i++){
            if(digits[i] == 0){
                dp.push_back(dp[i-1]);
            } else if(digits[i-1] == 0){
                dp.push_back(dp[i-1]+ fib[i]);
            } else{
                dp.push_back(fib[i-1] + fib[i]);
            }
            fib.push_back(fib[i] + fib[i-1]);
        }
        return dp[dp.size()-1];
    }
};
9000