600. Non-negative Integers without Consecutive Ones

Given a positive integer n, find the number of non-negative integers less than or equal to n, whose binary representations do NOT contain consecutive ones.

Example 1:

Input: 5
Output: 5
Explanation: 
Here are the non-negative integers <= 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

Note: 1 <= n <= 109

思路：
给一个数n，找出有多少个不大于n的正数，且这些正数的二进制表示中不包含两个连续的1。
int转2进制string的API

先用dp找出在k位二进制长度的数有多少满足条件的。a[i]表示以0为结尾的长度为i + 1的满足条件的数字个数，b[i]表示1为结尾的。

再找a[n - 1] + b[n - 1]里面overcount了多少，只有当连续两位为0时，b[i]是overcount的。11、01、10都能cover到a[i]和b[i]

class Solution
    public int findIntegers(int num) {
        StringBuilder sb = new StringBuilder(Integer.toBinaryString(num)).reverse();  
        int n = sb.length();  
        int[] a = new int[n];  
        int[] b = new int[n];  
        a[0] = 1;  
        b[0] = 1;  
        for (int i = 1; i < n; i++) {  
            a[i] = a[i - 1] + b[i - 1];  
            b[i] = a[i - 1];  
        }  
        int res = a[n - 1] + b[n - 1];  
        for (int i = n - 2; i >= 0; i--) {  
            if (sb.charAt(i) == '0' && sb.charAt(i + 1) == '0') {  
                res -= b[i];  
            }  
            if (sb.charAt(i) == '1' && sb.charAt(i + 1) == '1') {  
                break;  
            }  
        }  
        return