[Swift]LeetCode600. 不含連續1的非負整數 | 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

---

給定一個正整數 n，找出小於或等於 n 的非負整數中，其二進制表示不包含 連續的1 的個數。

示例 1:

輸入: 5
輸出: 5
解釋: 
下面是帶有相應二進制表示的非負整數<= 5：
0 : 0
1 : 1
2 : 10
3 : 11
4 : 100
5 : 101
其中，只有整數3違反規則（有兩個連續的1），其餘5個知足規則。

說明: 1 <= n <= 10^9

---

Runtime: 8 ms

Memory Usage: 18.5 MB

 1 class Solution {
 2     func findIntegers(_ num: Int) -> Int {
 3         var res:Int = 0
 4         var k:Int = 31
 5         var pre:Int = 0
 6         var f:[Int] = [Int](repeating:0,count:32)
 7         f[0] = 1
 8         f[1] = 2
 9         for i in 2..<31
10         {
11             f[i] = f[i - 2] + f[i - 1]
12         }
13         while (k >= 0)
14         {
15             if num & (1 << k) != 0
16             {
17                 res += f[k]
18                 if pre != 0
19                 {
20                     return res
21                 }
22                 pre = 1
23             }
24             else
25             {
26                 pre = 0
27             }
28             k -= 1
29         }
30         return res + 1
31     }
32 }