[Swift]LeetCode600. 不含连续1的非负整数 | Non-negative Integers without Consecutive Ones

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Given a positive integer n, find the number of non-negativeintegers 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 <= 10^9

---

给定一个正整数 n，找出小于或等于 n 的非负整数中，其二进制表示不包含 连续的1 的个数。

示例 1:

输入: 5
输出: 5
解释: 
下面是带有相应二进制表示的非负整数<= 5：
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
其中，只有整数3违反规则（有两个连续的1），其他5个满足规则。

说明: 1 <= n <= 10^9

---

Runtime: 8 ms

Memory Usage: 18.5 MB

class Solution {
    func findIntegers(_ num: Int) -> Int {
        var res:Int = 0
        var k:Int = 31
        var pre:Int = 0
        var f:[Int] = [Int](repeating:0,count:32)
        f[0] = 1
        f[1] = 2
        for i in 2..<31
        {
            f[i] = f[i - 2] + f[i - 1]
        }
        while (k >= 0)
        {
            if num & (1 << k) != 0
            {
                res += f[k]
                if pre != 0
                {
                    return res
                }
                pre = 1
            }
            else
            {
                pre = 0
            }
            k -= 1
        }
        return res + 1
    }
}