理论基础

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509. 斐波那契数

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视频讲解：手把手带你入门动态规划 | LeetCode：509.斐波那契数_哔哩哔哩_bilibili
普通递归：

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

动态规划：

class Solution {
    public int fib(int n) {
        if (n < 2) {
            return n;
        }
        int a = 0, b = 1, sum = 0;
        for (int i = 2; i <= n; i++) {
            sum = a + b;
            a = b;
            b = sum;
        }
        return sum;
    }
}

70. 爬楼梯

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dp 数组：

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }
        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}

状态压缩：

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }

        int first = 1;  // dp[i-2]
        int second = 2; // dp[i-1]
        int current = 0; // dp[i]

        for (int i = 3; i <= n; i++) {
            current = first + second;
            first = second;
            second = current;
        }

        return second; // dp[n]
    }
}

746. 使用最小花费爬楼梯

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视频讲解：动态规划开更了！| LeetCode：746. 使用最小花费爬楼梯_哔哩哔哩_bilibili

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        if (n <= 2) {
            return Math.min(cost[0], cost[1]);
        }
        int first = 0;
        int second = 0;
        int current = 0;
        for (int i = 2; i <= n; i++) {
            current = Math.min(first + cost[i - 2], second + cost[i - 1]);
            first = second;
            second = current;
        }
        return current;
    }
}