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
func findIntegers(n int) int {
	m := bits.Len(uint(n))
	f := make([][2]int, m)
	for i := range f {
		f[i] = [2]int{-1, -1}
	}
	var dfs func(i, pre int, limit bool) int
	dfs = func(i, pre int, limit bool) int {
		if i < 0 {
			return 1
		}
		if !limit && f[i][pre] != -1 {
			return f[i][pre]
		}
		up := 1
		if limit {
			up = n >> i & 1
		}
		ans := 0
		for 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
	}
	return dfs(m-1, 0, true)
}
