Leetcode 刷题笔记1 动态规划part01

leetcode 509 斐波那契数

递归：

class Solution:
    def fib(self, n: int) -> int:
        if n < 2:
            return n
        return self.fib(n - 1) + self.fib(n - 2)

动态规划：

class Solution:
    def fib(self, n: int) -> int:
        if n == 0:
            return 0
        dp = [0, 1]
        for i in range(2, n + 1):
            total = dp[0] + dp[1]
            dp[0] = dp[1]
            dp[1] = total
        return dp[1]

leetcode 70 爬楼梯

难点在于分析dp数组的递推公式

class Solution:
    def climbStairs(self, n: int) -> int:
        if n <= 2:
            return n
        dp = [0, 1, 2]
        for i in range(3, n + 1):
            total = dp[1] + dp[2]
            dp[1] = dp[2]
            dp[2] = total
        return dp[2]

leetcode 746 使用最小花费爬楼梯

1. 确定dp数组以及其下标含义
2. 分析递推公式
3. 初始化dp数组
4. 确定遍历顺序
5. 举例推导dp数组

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        dp = [0, 0]
        for i in range(2, len(cost) + 1):
            total = min(dp[0] + cost[i - 2], dp[1] + cost[i - 1])
            dp[0] = dp[1]
            dp[1] = total
        return dp[1]