【每天一題】劍指 Offer 10- I. 斐波那契數列

寫一個函數，輸入 n ，求斐波那契（Fibonacci）數列的第 n 項。斐波那契數列的定義如下：
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
斐波那契數列由 0 和 1 開始，之後的斐波那契數就是由之前的兩數相加而得出。
答案需要取模 1e9+7（1000000007），如計算初始結果爲：1000000008，請返回 1。

示例 1：
輸入：n = 2
輸出：1

示例 2：
輸入：n = 5
輸出：5

動態規劃：
原理： 以斐波那契數列性質 f(n + 1) = f(n) + f(n - 1)爲轉移方程。
從計算效率、空間複雜度上看，動態規劃是本題的最佳解法。

class Solution {
    public int fib(int n) {
        if(n < 0) return -1;
        if(n == 0) return 0;
        if(n == 1) return 1;
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        for(int i = 2; i <= n; i++){
            dp[i] = dp[i-1] + dp[i-2];
            dp[i] %= 1000000007;
        }
        return dp[n];
    }
}

循環求餘法：

class Solution {
    public int fib(int n) {
        int a = 0, b = 1, sum;
        for(int i = 0; i < n; i++){
            sum = (a + b) % 1000000007;
            a = b;
            b = sum;
        }
        return a;
    }
}

暴力樂呵樂呵算法

class Solution {
    int f[]={0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,134903163,836311896,971215059,807526948,778742000,586268941,365010934,951279875,316290802,267570670,583861472,851432142,435293607,286725742,722019349,8745084,730764433,739509517,470273943,209783453,680057396,889840849,569898238,459739080,29637311,489376391,519013702,8390086,527403788,535793874,63197655,598991529,662189184,261180706,923369890,184550589,107920472,292471061,400391533,692862594,93254120,786116714,879370834,665487541,544858368,210345902,755204270,965550172,720754435,686304600,407059028,93363621,500422649,593786270,94208912,687995182};
    int fib(int n){
        return f[n];
    }
}