600. Non-negative Integers without Consecutive Ones

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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 rule. 

Note: 1 <= n <= 109

Integer.toBinaryString(num);得到的string的index为0出的char一定是1

/*
 * 先考虑简单的情况：n位任意取，可用DP
 * 接着考虑有限制的情况：第一位取0，那么后面的n-1位任意取；第一位取1，那第二位只能去0，但是
 * 如果第二位原来是1，那此时后面n-2位任意取，如果原来是0，那就递归调用findIntegers函数
 */
public class Solution {
    public int findIntegers(int num) {
    	if(num == 0)	return 1;
    	if(num == 1)	return 2;
    	if(num == 2)	return 3;
    	if(num == 3)	return 3;
        String s = Integer.toBinaryString(num);
        int n = s.length();
        
        int[][] dp = new int[n+1][2];
        dp[1][0] = 1;	dp[1][1] = 1;
        for(int i=2; i<=n; i++) {
        	dp[i][0] = dp[i-1][0]+dp[i-1][1];
        	dp[i][1] = dp[i-1][0];
        }
        
        if(s.charAt(1) == '1')
        	return dp[n-1][0] + dp[n-1][1] + dp[n-2][0] + dp[n-2][1];
        else
        	return dp[n-1][0] + dp[n-1][1] + findIntegers(Integer.valueOf(s.substring(2), 2));
    }
}