代码随想录算法训练营第38天| 509. 斐波那契数、70. 爬楼梯、746. 使用最小花费爬楼梯

509. 斐波那契数:


代码思路
dp[i]代表F(i),然后前面的推后面的,也就是所谓状态转移方程(题目已给出)

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

70. 爬楼梯:


代码思路
和斐波那契一模一样

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

746. 使用最小花费爬楼梯:


代码思路
固定套路,一阶动规。

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

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