代码随想录算法训练营第36天 | 动态规划基础2:62.不同路径、63.不同路径 II

62.不同路径
https://leetcode.cn/problems/unique-paths/submissions/548656029/
代码随想录
https://programmercarl.com/0062.不同路径.html
63.不同路径 II
https://leetcode.cn/problems/unique-paths-ii/description/
代码随想录
https://programmercarl.com/0063.不同路径II.html#算法公开课
343.整数拆分（一刷跳过）
https://leetcode.cn/problems/integer-break/
代码随想录
https://programmercarl.com/0343.整数拆分.html
96.不同的二叉搜索树（一刷跳过）
https://leetcode.cn/problems/unique-binary-search-trees/description/
代码随想录
https://programmercarl.com/0096.不同的二叉搜索树.html

62.不同路径

题解思路

• 思路一：动态规划算法：
  • dp[i][j]代表到第i行第j列的格子的可能性；
  • dp[i][j] = dp[i-1][j]+dp[i][j-1]
• 排列组合计算
  -计算 $ C(m+n-2,m-1) $

题解代码

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[1]*n for _ in range(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[-1][-1]

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        count = m-1
        t = m+n-2
        fenmu = m-1
        fenzi = 1
        while count>0:
            fenzi*=t ##分子已知变化
            t-=1

            while fenmu!=0 and fenzi%fenmu==0:###除得尽再除 
                fenzi //= fenmu
                fenmu -=1
            count-=1
        return fenzi

63.不同路径 II

题解思路

• 和一比只需要多处理几个关于障碍的特殊情况
  • 障碍在起点和终点 直接为0
  • 初始化时，如果出现障碍，剩下的全为0

题解代码

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m =  len(obstacleGrid)
        n = len(obstacleGrid[0])

        ## 绝路不计算
        if obstacleGrid[0][0]==1 or obstacleGrid[-1][-1]:
            return 0

        dp = [[0]*n for _ in range(m)]

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

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

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