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] = 1

        for j in range(n):
            dp[0][j] = 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]

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 0

        for i in range(m):
            if obstacleGrid[i][0]==0:
                dp[i][0] = 1
            else:
                break
        
        for j in range(n):
            if obstacleGrid[0][j]==0:
                dp[0][j] = 1
            else:
                break

        for 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] = 1
        
        for 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]