


Detailed example of Python using the backtracking method subset tree template to solve the stair climbing problem
This article mainly introduces Python's use of the backtracking method subset tree template to solve the stair climbing problem. It briefly explains the stair climbing problem and combines it with examples to provide relevant operating skills for Python's backtracking method subset tree template to solve the stair climbing problem. It is required Friends can refer to
. This article describes an example of Python using the backtracking method subset tree template to solve the stair climbing problem. Share it with everyone for your reference. The details are as follows:
Question
A certain staircase has n steps, and each step can only take 1 step. , or 2 steps. How many ways are there to climb stairs from bottom to top?
Analysis
This problem has been solved before using the divide and conquer method. However, here I am going to use the backtracking subset tree template to solve it.
Bring out the element-state space analysis method: each step is an element, and the number of steps [1,2] that can be taken is its state space. It is not difficult to see that the elements are not fixed, but the state space is fixed.
Upload the code directly.
Code
'''爬楼梯''' n = 7 # 楼梯阶数 x = [] # 一个解(长度不固定,1-2数组,表示该步走的台阶数) X = [] # 一组解 # 冲突检测 def conflict(k): global n, x, X # 部分解步的步数之和超过总台阶数 if sum(x[:k+1]) > n: return True return False # 无冲突 # 回溯法(递归版本) def climb_stairs(k): # 走第k步 global n, x, X if sum(x) == n: # 已走的所有步数之和等于楼梯总台阶数 print(x) #X.append(x[:]) # 保存(一个解) else: for i in [1, 2]: # 第k步这个元素的状态空间为[1,2] x.append(i) if not conflict(k): # 剪枝 climb_stairs(k+1) x.pop() # 回溯 # 测试 climb_stairs(0) # 走第0步
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