Given the number k
, return the minimum number of Fibonacci numbers whose sum is equal to k
, whether a Fibonacci number could be used multiple times.
The Fibonacci numbers are defined as:
- F1 = 1
- F2 = 1
- Fn = Fn-1 + Fn-2 , for n > 2.
It is guaranteed that for the given constraints we can always find such fibonacci numbers that sum k
.
Example 1:
Input: k = 7 Output: 2 Explanation: The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ... For k = 7 we can use 2 + 5 = 7.
Example 2:
Input: k = 10 Output: 2 Explanation: For k = 10 we can use 2 + 8 = 10.
Example 3:
Input: k = 19 Output: 3 Explanation: For k = 19 we can use 1 + 5 + 13 = 19.
Constraints:
1 <= k <= 10^9
class Solution { public int findMinFibonacciNumbers(int k) { TreeSet<Integer> set = new TreeSet(); int a = 0, b = 1; int c = a + b; set.add(1); while(c <= k) { c = a + b; set.add(c); a = b; b = c; } int res = 0; while(k > 0) { int cur = set.floor(k); res++; k -= cur; } return res; } }
Fibonacci数的产生要三个变量,记一下
这题用TreeSet的floor方法,返回离k最近(小于等于)的key,然后继续往下即可.类似的还有ceiling方法,返回大于等于的key