题目一

上边、左边初始化为1

采用求和进行dp运算

class Solution(object):def uniquePaths(self, m, n):""":type m: int:type n: int:rtype: int"""dp =[[0]*n for _ in range(m)]for i in range(m):dp[i][0] = 1for j in range(n):dp[0][j] = 1for 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]

第二题

上题变种，

初始化策略要改变，遇到障碍后，之后的都是0

dp更新策略改变 ，遇到障碍跳过

class Solution(object):def uniquePathsWithObstacles(self, obstacleGrid):""":type obstacleGrid: List[List[int]]:rtype: int"""m = len(obstacleGrid)n = len(obstacleGrid[0])dp = [[0] * n for _ in range(m)]if obstacleGrid[0][0]==1 or obstacleGrid[m-1][n-1]==1:return 0for i in range(m):if obstacleGrid[i][0]==0:dp[i][0] = 1else:breakfor j in range(n):if obstacleGrid[0][j]==0:dp[0][j] = 1else:breakfor i in range(1,m):for j in range(1,n):if obstacleGrid[i][j] == 0:dp[i][j] = dp[i-1][j] + dp[i][j-1]return dp[m-1][n-1]

题目三：

dp[i] 代表将 i 拆分后的乘积最大值，

计算方式有两种：j x i-j

或者  j x dp[i-j]

本质上也是二元的思维，

class Solution(object):def integerBreak(self, n):""":type n: int:rtype: int"""dp = [0] * (n+1)dp[2] = 1for i in range(3,n+1):for j in range(1,i):dp[i] = max(dp[i],j*(i-j),j*dp[i-j])return dp[n]