Datawhale编程实践（LeetCode 腾讯精选练习50）Task8

1.不同路径https://leetcode-cn.com/problems/unique-paths/

看到这题，我第一想到的也是组合数学。就直接计算一下组合数就行了

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        return comb(m + n - 2, n - 1)

在官方解答看到了动态规划的解法，也学习了一下动态规划的思想。当前状态为前一个状态的和，此处的前一个状态有2种(从左边走过来，从上边走过来)

所以状态转移方程为f(i, j) = f(i-1, j) + f(i, j-1)

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        f = [[1] * n] + [[1] + [0] * (n - 1) for _ in range(m - 1)]
        print(f)
        for i in range(1, m):
            for j in range(1, n):
                f[i][j] = f[i - 1][j] + f[i][j - 1]
        return f[m - 1][n - 1]

这里

f = [[1] * n] + [[1] + [0] * (n - 1) for _ in range(m - 1)]

把我惊艳了一下，又学到了新的python的语法，如果是我应该是傻傻地用for循环去创建吧

这里通过for _ in range(m - 1)避免浅复制问题https://blog.csdn.net/qq_21997625/article/details/105694880

复习了一下python list运算

https://blog.csdn.net/jp_666/article/details/98055191

https://www.runoob.com/python3/python3-list-operator.html

2.爬楼梯https://leetcode-cn.com/problems/climbing-stairs/

3.子集https://leetcode-cn.com/problems/subsets/