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[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
 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 }

 

posted @ 2019-02-28 16:48  为敢技术  阅读(300)  评论(0编辑  收藏  举报