509. 斐波那契数

class Solution {
    public int fib(int n) {
        // base case
        if(n == 0)return 0;
        if(n == 1)return 1;
        
        int[] f = new int[n + 1];

        f[0] = 0;
        f[1] = 1;

        for(int i = 2; i <= n; i++){
            f[i] = f[i-1] + f[i-2];
        }

        return f[n];
    }
}

70. 爬楼梯

class Solution {
    public int climbStairs(int n) {
        // dp[n]: 爬到第n阶的方法数
        int[] dp = new int[n + 1];

        dp[0] = 1;
        dp[1] = 1;

        if(n == 0 || n == 1)return 1;

        // 爬到第i阶的方法数等于i-1阶和i-2阶的方法数之和
        for(int i = 2; i <= n; i++){
            dp[i] = dp[i - 1] + dp[i - 2];
        }

        return dp[n];
    }
}

746. 使用最小花费爬楼梯

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        // dp[n]: 达到第n阶时的最低花费
        int[] dp = new int[n + 1];

        dp[0] = 0;
        dp[1] = 0;

        for(int i = 2; i <= n; i++){
            dp[i] = Math.min((dp[i - 1] + cost[i - 1]), (dp[i - 2]+ cost[i - 2]));
        }

        return dp[n];
    }
}