题解 | #跳台阶#

动态规划解法：

    public int jumpFloor(int target) {
        //定义dp[n]为跳到第n层的有几种跳法
        //且dp[n] = dp[n-1] + dp[n-2];
        if (target <= 1) return 1;

        int[] dp = new int[target + 1];
        dp[0] = 1;
        dp[1] = 1;
        //控制循环次数和dp数组索引
        for (int i = 2; i <= target; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[target];
    }