算法修炼Day38|509. 斐波那契数 ● 70. 爬楼梯 ● 746. 使用最小花费爬楼梯

40 阅读1分钟
题目:509. 斐波那契数 - 力扣(LeetCode)
代码实现:
class Solution {
    public int fib(int n) {
        if (n <= 1) return n; // 必须加上避免越界
        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];
        }
        return dp[n];
    }
}
题目:70. 爬楼梯 - 力扣(LeetCode)
代码实现:
class Solution {
    public int climbStairs(int n) {
        if (n <= 2) return n;
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}
题目:746. 使用最小花费爬楼梯 - 力扣(LeetCode)
代码实现:
class Solution {
    public int minCostClimbingStairs(int[] cost) {
        // dp[i] 表示到达i位置支付的最小花费
        int[] dp = new int[cost.length];
        dp[0] = cost[0];
        dp[1] = cost[1];
        for (int i = 2; i < cost.length; i++) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return Math.min(dp[cost.length - 1], dp[cost.length - 2]);
    }
}