一、题目描述
示例 1:
输入:n = 2
输出:1
解释:F(2) = F(1) + F(0) = 1 + 0 = 1
示例 2:
输入:n = 3
输出:2
解释:F(3) = F(2) + F(1) = 1 + 1 = 2
示例 3:
输入:n = 4
输出:3
解释:F(4) = F(3) + F(2) = 2 + 1 = 3
Leetcode链接: 509. 斐波那契数
二、参考代码
//1
class Solution {
public:
int fib(int n) {
if( n == 0)
{
return 0;
}
if(n <= 2 && n > 0)
{
return 1;
}
vector<int>dp(n+1);
dp[0] = 0;
dp[1] = 1;
for(int i=2;i<=n;i++)
{
dp[i] = dp[i-1]+dp[i-2];
}
return dp[n];
}
};
//2
class Solution {
public:
int fib(int n) {
if( n == 0)
{
return 0;
}
if(n <= 2 && n > 0)
{
return 1;
}
else
{
return fib(n-1) + fib(n-2);
}
}
};
//3
class Solution {
public:
int fib(int n) {
if( n == 0)
{
return 0;
}
if(n <= 2 && n > 0)
{
return 1;
}
else
{
int p =0,q=0,r=1;
for(int i=2;i<=n;i++)
{
p = q;
q = r;
r = p + q;
}
return r;
}
}
};