function findSwappedNodes(root) inorderArray = [] inorderTraversal(root, inorderArray) x = y = -1 for i in range(1, len(inorderArray)): if inorderArray[i] < inorderArray[i - 1]: y = inorderArray[i] if x == -1: x = inorderArray[i - 1] else: break a, b = findNodes(root, x, y) swap(a, b):分析代码并且计算运行次数
时间: 2023-02-25 14:11:00 浏览: 38
这段代码实现了某个二叉搜索树中两个节点的交换,首先它调用inorderTraversal函数将树的中序遍历结果存入inorderArray数组,然后通过for循环从1开始遍历inorderArray数组,当当前数小于前一个数时,表示两个节点位置交换了,将当前数存入y,如果x还没有被赋值,则将前一个数存入x,否则break跳出循环,最后调用findNodes函数获取交换节点的两个节点,并调用swap函数交换这两个节点。
计算运行次数方面,具体要看findNodes和swap函数的实现,以及二叉搜索树的大小,总的时间复杂度可以认为是O(N)的。
相关问题
Solve the following differential equation using symbolic and numeric methods: y'' + 2y' + y = 2sin(x), y(0) = 0, y'(0) = 1
Symbolic method:
We can solve the differential equation using the characteristic equation:
r^2 + 2r + 1 = 0
Solving for r, we get:
r = -1 (repeated root)
So the general solution is:
y(x) = (c1 + c2*x)*e^(-x)
To find the particular solution, we can use the method of undetermined coefficients. Since the right-hand side of the equation is 2sin(x), we can guess a particular solution of the form:
y_p(x) = A*sin(x) + B*cos(x)
Taking the first and second derivatives of y_p(x), we get:
y'_p(x) = A*cos(x) - B*sin(x)
y''_p(x) = -A*sin(x) - B*cos(x)
Substituting these into the differential equation, we get:
-A*sin(x) - B*cos(x) + 2(A*cos(x) - B*sin(x)) + (A*sin(x) + B*cos(x)) = 2sin(x)
Simplifying, we get:
2A = 2
-2B + 2A = 0
Solving for A and B, we get:
A = 1
B = 1
So the particular solution is:
y_p(x) = sin(x) + cos(x)
The general solution is:
y(x) = (c1 + c2*x)*e^(-x) + sin(x) + cos(x)
Using the initial conditions, we get:
y(0) = c1 + 1 = 0
y'(0) = -c1 + c2 + 1 = 1
Solving for c1 and c2, we get:
c1 = -1
c2 = 2
So the solution to the differential equation is:
y(x) = (-1 + 2x)*e^(-x) + sin(x) + cos(x)
Numeric method:
We can also solve the differential equation numerically using a computer program. One way to do this is to use the fourth-order Runge-Kutta method. Here is an example Python code that solves the differential equation:
```python
import numpy as np
import matplotlib.pyplot as plt
def f(x, y):
return [y[1], 2*np.sin(x) - 2*y[1] - y[0]]
x0 = 0
y0 = [0, 1]
h = 0.01
x = np.arange(x0, 10+h, h)
y = np.zeros((len(x), len(y0)))
y[0] = y0
for i in range(len(x)-1):
k1 = h*np.array(f(x[i], y[i]))
k2 = h*np.array(f(x[i]+h/2, y[i]+k1/2))
k3 = h*np.array(f(x[i]+h/2, y[i]+k2/2))
k4 = h*np.array(f(x[i]+h, y[i]+k3))
y[i+1] = y[i] + (k1 + 2*k2 + 2*k3 + k4)/6
plt.plot(x, y[:,0])
plt.xlabel('x')
plt.ylabel('y')
plt.show()
```
The code defines the function f(x, y) that returns the right-hand side of the differential equation as a list. It then initializes the initial conditions and the step size. The program then uses the fourth-order Runge-Kutta method to solve the differential equation and stores the solution in the array y. Finally, the program plots the solution. The resulting plot should match the symbolic solution.
用python和canvas绘制-3 * x * x * x -3 * x *x + 4 * sin(x)
好的,我可以给你提供一个简单的Python和Canvas的绘制代码,来绘制函数-3 * x * x * x -3 * x *x + 4 * sin(x)的图形。
首先,你需要安装Python的matplotlib库,这个库可以支持在Python中进行绘图。你可以使用pip来安装这个库,命令如下:
```
pip install matplotlib
```
然后,你可以使用下面的Python代码来绘制这个函数的图形:
``` python
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-10, 10, 1000)
y = -3 * x * x * x - 3 * x * x + 4 * np.sin(x)
plt.plot(x, y)
plt.xlabel('x')
plt.ylabel('y')
plt.title('-3 * x * x * x -3 * x *x + 4 * sin(x)')
plt.show()
```
这个代码会在窗口中绘制出函数的图形。如果你想要使用Canvas进行绘制,你可以使用下面的代码:
``` python
from tkinter import *
import numpy as np
def f(x):
return -3 * x * x * x - 3 * x * x + 4 * np.sin(x)
root = Tk()
root.title("Function Plotter")
root.geometry("500x500")
canvas = Canvas(root, width=400, height=400, bg='white')
canvas.pack()
x0, y0 = 200, 200
scale = 20
x = np.linspace(-10, 10, 1000)
y = f(x)
points = []
for i in range(len(x)):
points.append((x0 + x[i] * scale, y0 - y[i] * scale))
canvas.create_line(points, fill='blue')
root.mainloop()
```
这个代码会在一个窗口中使用Canvas绘制出函数的图形。你可以调整窗口的大小和绘制的比例来获得更好的效果。
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