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一、題目描述

示例 1：

輸入：n = 2
輸出：1
解釋：F(2) = F(1) + F(0) = 1 + 0 = 1
示例 2：

輸入：n = 3
輸出：2
解釋：F(3) = F(2) + F(1) = 1 + 1 = 2
示例 3：

輸入：n = 4
輸出：3
解釋：F(4) = F(3) + F(2) = 2 + 1 = 3

Leetcode鏈接: 509. 斐波那契數

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二、參考代碼

//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;}}
};

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