青蛙跳台阶

递推公式：

和斐波那契数列是一致的

1.暴力办法，时间复杂度O(2^n)

public class Solution {
    public int JumpFloor(int n) {
        if (n == 1) return 1; 
        if (n == 2) return 2;
        return JumpFloor(n - 1) + JumpFloor(n - 2);
    }
}

2.动态规划

class Solution {
    static final int MODULO = 1000000007;

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