目录

学习目标

学习内容

 理论基础

 509. 斐波那契数 

70. 爬楼梯

 746. 使用最小花费爬楼梯

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学习目标

学习内容

 理论基础

problems/动态规划理论基础.md · programmercarl/leetcode-master（代码随想录出品） - Gitee.comhttps://gitee.com/programmercarl/leetcode-master/blob/master/problems/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%E7%90%86%E8%AE%BA%E5%9F%BA%E7%A1%80.md

 509. 斐波那契数 

509. 斐波那契数 - 力扣（LeetCode）https://leetcode.cn/problems/fibonacci-number/

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

class Solution:
    def fib(self, n: int) -> int:
        @cache
        def dfs(i):
            if i<2:return i
            return dfs(i-1)+dfs(i-2)
        return dfs(n)

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

class Solution:
    def fib(self, n: int) -> int:
        if n<2:return n
        a = 0
        b = 1
        for i in range(2,n+1):
            a,b=b,a+b
        return b

70. 爬楼梯

70. 爬楼梯 - 力扣（LeetCode）https://leetcode.cn/problems/climbing-stairs/

class Solution:
    def climbStairs(self, n: int) -> int:
        @cache
        def dfs(i):
            if i<2:return 1
            return dfs(i-1)+dfs(i-2)
        return dfs(n)

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

 746. 使用最小花费爬楼梯

746. 使用最小花费爬楼梯 - 力扣（LeetCode）https://leetcode.cn/problems/min-cost-climbing-stairs/

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        @cache
        def dfs(i):
            if i<2:return 0
            return min(dfs(i-1)+cost[i-1],dfs(i-2)+cost[i-2])
        return dfs(len(cost))

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