Non-negative Integers without Consecutive Ones
<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>
Dynamic Programming
class Solution {
    private int n;
    private Integer[][] f;

    public int findIntegers(int n) {
        this.n = n;
        int m = Integer.SIZE - Integer.numberOfLeadingZeros(n);
        f = new Integer[m][2];
        return dfs(m - 1, 0, true);
    }

    private int dfs(int i, int pre, boolean limit) {
        if (i < 0) {
            return 1;
        }
        if (!limit && f[i][pre] != null) {
            return f[i][pre];
        }
        int up = limit ? (n >> i & 1) : 1;
        int ans = 0;
        for (int j = 0; j <= up; ++j) {
            if (j == 1 && pre == 1) {
                continue;
            }
            ans += dfs(i - 1, j, limit && j == up);
        }
        if (!limit) {
            f[i][pre] = ans;
        }
        return ans;
    }
}
