509. Fibonacci Number
Difficulty: 简单
The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,
F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.
Given n, calculate F(n).
Example 1:
Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Example 2:
Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Example 3:
Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
Constraints:
0 <= n <= 30
Solution 1
Language: c++
class Solution {
public:
int fib(int n) {
if(n == 0 || n == 1)
return n;
else
return fib(n-1)+fib(n-2);
}
};
Language: python
class Solution:
def fib(self, n: int) -> int:
if n == 1 or n == 0:
return n
else:
return self.fib(n-1) + self.fib(n-2)
使用递归的，重复计算了很多次，不过考虑到不超过30次，其实还能接受，但是如果能记录下重复计算的结果，使用空间换时间的思路。
Solution 2
Language: c++
class Solution {
public:
int fib(int n) {
if (n == 0 || n == 1)
return n;
int a = 0,b = 1;
for(int i = 2;i<=n;++i){
b = a + b;
a = b - a;
}
return b;
}
};
Language: python
class Solution:
def fib(self, n: int) -> int:
if n == 1 or n == 0:
return n
a,b=0,1
for i in range(1,n):
b = a + b
a = b - a
return b
去除掉重复的计算，速度也会快不少，使用的空间也只有O(1)。
最后贴出出lc的运行消耗对比
