斐波那契数列

解法一：

public class Solution {
    public int Fibonacci(int n) {
        if (n <= 1) return n;
        int a0 = 0, a1 = 1, ret = 0;
        for (int i = 2; i <= n; i++) {
            ret = a0 + a1;
            a0 = a1;
            a1 = ret;
        }
        return ret;
    }
}

解法二：

进一步减少存储变量个数到2个：

public class Solution {
    public int Fibonacci(int n) {
        if (n <= 1) return n;
        int one = 0, sum = 1;
        for(int i = 2; i <= n; i++) {
            sum = sum + one;
            one = sum - one;
        }
        return sum;
    }
}