【代码随想录训练营】【Day 41】【动态规划-1 and 2】| Leetcode 509, 70, 746, 62, 63

【代码随想录训练营】【Day 41】【动态规划-1 and 2】| Leetcode 509, 70, 746, 62, 63

需强化知识点

题目

509. 斐波那契数

class Solution:
    def fib(self, n: int) -> int:
        if n == 0:
            return 0
        if n == 1:
            return 1
        dp = [0] * (n+1)
        dp[0], dp[1] = 0, 1

        for i in range(2, n+1):
            dp[i] = dp[i-1] + dp[i-2]
        
        return dp[n]
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70. 爬楼梯

class Solution:
    def climbStairs(self, n: int) -> int:
        if n == 1:
            return 1
        if n == 2:
            return 2
        dp = [1] * n
        dp[0], dp[1] = 1, 2

        for i in range(2, n):
            dp[i] = dp[i-1] + dp[i-2]
        
        return dp[n-1]

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746. 使用最小花费爬楼梯

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:

        # 爬上台阶 i 所需支付的费用
        dp = [0] * (len(cost)+1)
        
        for i in range(2, len(cost)+1):
            dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2])
        
        return dp[len(cost)]          
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62. 不同路径

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0] * n for _ in range(m)]

        for i in range(m):
            dp[i][0] = 1
        for i in range(n):
            dp[0][i] = 1
        
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
        
        return dp[m-1][n-1]       
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63. 不同路径 II

• 注意初始化和遍历时，遇到障碍物的处理

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for _ in range(m)]

        for i in range(m):
            if obstacleGrid[i][0] == 1:
                break
            dp[i][0] = 1
        
        for i in range(n):
            if obstacleGrid[0][i] == 1:
                break
            dp[0][i] = 1
        
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
        
        return dp[m-1][n-1]         
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