LeetCode-----フィボナッチ数列
, n , (Fibonacci) n 。 :
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), N > 1.
0 1 , 。
1e9+7(1000000007), :1000000008, 1。
: (LeetCode)
:https://leetcode-cn.com/problems/fei-bo-na-qi-shu-lie-lcof
。 , 。
class Solution {
public:
int fib(int n) {
if (n <= 1)
{
return n;
}
else
{
return fib(n - 1) + fib(n - 2);
}
}
};
*解法2:
class Solution {
public:
int fib(int n) {
if (n == 0 || n == 1)
return n;
int a = 1, b = 0;
for (int i = 1; i < n; i++) {
a = a + b;
b = a - b;
a %= 1000000007;
}
return a;
}
};