【代码随想录】动态规划专题1（Python）_代码随想录 动态规划 pdf

1.斐波那契数

509. 斐波那契数
dp[i] = dp[i-1] + dp[i-2]

class Solution:
    def fib(self, n: int) -> int:
        if n==0 : return 0
        dp0 = 1
        dp1 = 1
        for i in range(2,n):
            r = dp0 + dp1
            dp0 = dp1
            dp1 = r
        return dp1
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2.爬楼梯

70. 爬楼梯
同1

class Solution:
    def climbStairs(self, n: int) -> int:
        dp0 = 1
        dp1 = 1
        for i in range(2,n+1):
            r = dp0 + dp1
            dp0 = dp1
            dp1 = r
        return dp1
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3.使用最小花费爬楼梯

746. 使用最小花费爬楼梯
dp[i]表示到第i阶的最小花费
dp[i] = min( dp[i-1]+cost[i-1], dp[i-2]+cost[i-2])
dp0 = dp1 = 0

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

62.不同路径
每个格子只能由上格和左格走到
dp[i][j] = dp[i-1][j] + dp[i][j-1]

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[1]*n]*m
        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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5.不同路径 II

63. 不同路径 II
初始化时，首行首列遇到障碍物后续都为0，其余部分障碍物为0

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
            if obstacleGrid[0][0]==1: return 0
            for i in range(len(obstacleGrid)):
                if obstacleGrid[i][0] == 0:
                    obstacleGrid[i][0] = 1
                else:
                    for j in range(i,len(obstacleGrid)):
                        obstacleGrid[j][0] = 0
                    break
            for i in range(1,len(obstacleGrid[0])):
                if obstacleGrid[0][i] == 0:
                    obstacleGrid[0][i] = 1
                else:
                    for j in range(i,len(obstacleGrid[0])):
                        obstacleGrid[0][j] = 0
                    break
            for i in range(1,len(obstacleGrid)):
                for j in range(1,len(obstacleGrid[0])):
                    if obstacleGrid[i][j]==0:
                        obstacleGrid[i][j] = 1
                    else:
                        obstacleGrid[i][j] = 0
            for i in range(1,len(obstacleGrid)):
                for j in range(1,len(obstacleGrid[0])):
                    if obstacleGrid[i][j]!=0:
                        obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
            return obstacleGrid[-1][-1]
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6.整数拆分

343. 整数拆分
dp[i]表示拆分i可得的最大乘积
dp[i] = max( j*(i-j), j*dp[i-j], dp[i])
dp[0]=dp[1]=1

class Solution:
    def integerBreak(self, n: int) -> int:
        dp = [1]*(n+1)
        for i in range(2,n+1):
            for j in range(1,i):
                dp[i] = max( j*(i-j), j*dp[i-j], dp[i] )
        return dp[-1]
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7.不同的二叉搜索树

96.不同的二叉搜索树
dp[i]表示i个节点的二叉树的构成方式总数
dp[i] = dp[左树]*dp[右树] 的和
例如dp[3]=dp[0]*dp[2]+dp[1]*dp[1]+dp[2]*dp[0]

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