DP | 凛冬之怒

# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def rob(self, root: TreeNode) -> int: def superrob(node): # returns tuple of size two (now, later) # now: max money earned if input node is robbed # later: max money earned if input node is not robbed # base case if not node: return (0, 0) # get values left, right = superrob(node.left), superrob(node.right) # rob now now = node.val + left[1] + right[1] # rob later later = max(left) + max(right) return (now, later) return max(superrob(root))

这两天鱼神要去面Google，居然破天荒的准备了起来。昨天晚上他突然给我发了一道Leetcode上的Easy题 —— Climbing Stairs。我心想按照鱼神的水准，这种题怎么可能入得了他的法眼。后来看了看其实这道题还是挺能看出一个人的水平的（突然想起了求素数）这道题由于比较基础，所以解法相对来说也是比较多的。首先先上题目描述：

You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Note: Given n will be a positive integer.

因为在学习动态规划的时候，这道题就是用来入门的，所以一拿到这道题，应该很自然的就可以想到动态规划的方法。状态转移方程为：ways[n] = ways[n – 1] + ways[n – 2] 然后套路的发现这个形式长的居然和斐波那契数列一摸一样（对！没错，这题就是套着斐波那契出的。）所以就产生下面的几种解法：
Method 1: 递归求解

class Solution(object):
    def climbStairs(self, n):
        if n == 1 or n <= 0:
            return 1
        return climbStairs(n-1) + climbStairs(n-2)

Method 2: 非递归求解

class Solution(object):
    def climbStairs(self, n):
        a = b = 1
        for x in range(2, n + 1):
            a, b = b, a + b 
        return b

Method 3：通项公式求解

class Solution(object):
    def climbStairs(self, n):
        sqrt5 = math.sqrt(5)
        Phi = (1 + sqrt5) / 2
        phi = (1 - sqrt5) / 2
        return int((Phi ** (n + 1) - phi ** (n + 1)) / sqrt5)

Method 4: 杨辉三角叠加法

class Solution(object):
    def c(self, m, n):
        x = 1
        for i in xrange(m-n+1, m+1):
            x *= i
        for i in xrange(2, n+1):
            x /= i
        return x
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        result = 0
        for i in range(0, n/2+1):
            result += self.c(n-i, i)
        return result

$F(x) = \sum_{i = 1}^{\left \lfloor \frac{n}{2} \right \rfloor}C_{n-i}^{i}$

凛冬之怒

无为善道止战非攻

Tag Archives: DP

House Robber III

由爬楼梯问题引发的思考