必须要会的数据结构

实现一个二分查找

# 实现一个二分查找
# 输入:一个顺序list
# 输出: 待查找的元素的位置
def binarySearch(alist, item):
    first = 0
    last = len(alist) - 1

    while first <= last:
        mid = (first + last)//2
        print(mid)
        if alist[mid] > item:
            last = mid - 1
        elif alist[mid] < item:
            first = mid + 1
        else:
            return mid+1
    return -1

test = [0, 1, 2, 8, 13, 17, 19, 32, 42]
print(binarySearch(test, 3))


#4
#1
#2
#3
#-1

定义一个二叉树

class BinaryTree:
    def __init__(self, rootObj):
        self.key = rootObj
        self.leftChild = None
        self.rightChild = None

    def insertLeft(self, newNode):
        if self.leftChild == None:
            self.leftChild = BinaryTree(newNode)
        else:
            t = BinaryTree(newNode)
            t.leftChild = self.leftChild
            self.leftChild = t

    def insertRight(self, newNode):
        if self.rightChild == None:
            self.rightChild = BinaryTree(newNode)
        else:
            t = BinaryTree(newNode)
            t.rightChild = self.rightChild
            self.rightChild = t

    def getRightChild(self):
        return self.rightChild

    def getLeftChild(self):
        return self.leftChild

    def setRootVal(self, obj):
        self.key = obj

    def getRootVal(self):
        return self.key

    # 树的前序遍历
    # 树的后序遍历以及中序遍历见ParseTree.py
    def preorder(self):
        print(self.key)
        if self.leftChild:
            self.leftChild.preorder()
        if self.rightChild:
            self.rightChild.preorder()


# r = BinaryTree('a')
# print(r.getRootVal())
# print(r.getLeftChild())
# r.insertLeft('b')
# print(r.getLeftChild())
# print(r.getLeftChild().getRootVal())
# r.insertRight('c')
# print(r.getRightChild())
# print(r.getRightChild().getRootVal())
# r.getRightChild().setRootVal('hello')
# print(r.getRightChild().getRootVal())
a
None
<__main__.BinaryTree object at 0x0000029C419DCC88>
b
<__main__.BinaryTree object at 0x0000029C419DCEB8>
c
hello

选择排序

# 选择排序
def selectionSort(alist):
    for i in range(len(alist)-1):
        min = i
        for j in range(i+1, len(alist)):
            if alist[j] < alist[min]:
                min = j
        alist[i], alist[min] = alist[min], alist[i]
    return alist

alist = [54,26,93,17,77,31,44,55,20]
print(selectionSort(alist)) # [17, 20, 26, 31, 44, 54, 55, 77, 93]

实现冒泡排序

# Python 实现冒泡排序
def bubbleSort(alist):
    for passnum in range(len(alist)-1, 0, -1):
        for i in range(passnum):
            if alist[i] > alist[i+1]:
                alist[i], alist[i+1] = alist[i+1], alist[i]
    return alist

alist = [54,26,93,17,77,31,44,55,20]
print(bubbleSort(alist))

# 改进的冒泡排序, 加入一个校验, 如果某次循环发现没有发生数值交换, 直接跳出循环
def modiBubbleSort(alist):
    exchange = True
    passnum = len(alist) - 1
    while passnum >= 1 and exchange:
        exchange = False
        for i in range(passnum):
            if alist[i] > alist[i+1]:
                alist[i], alist[i+1] = alist[i+1], alist[i]
                exchange = True
        passnum -= 1
    return alist

print(bubbleSort(alist))


[17, 20, 26, 31, 44, 54, 55, 77, 93]
[17, 20, 26, 31, 44, 54, 55, 77, 93]

定义队列

class Queue:
    def __init__(self):
        self.items = []

    def isEmpty(self):
        return self.items == []

    def enqueue(self, item):
        self.items.insert(0, item)

    def dequeue(self):
        return self.items.pop()

    def size(self):
        return len(self.items)

q = Queue()
q.enqueue(4)
q.enqueue('god')
q.enqueue(True)
print(q.dequeue())

def hotPotato(namelist, num):
    simqueue = Queue()
    for name in namelist:
        simqueue.enqueue(name)

    while simqueue.size() > 1:
        for i in range(num):
            simqueue.enqueue(simqueue.dequeue())
        simqueue.dequeue()

    return simqueue.dequeue()

print(hotPotato(["Bill","David","Susan","Jane","Kent","Brad"],7))

# 设计一个良策都能插入删除的队列deque
class Deque:
    def __init__(self):
        self.items = []

    def isEmpty(self):
        return self.items == []

    def addRear(self, item):
        self.items.insert(0, item)

    def addFront(self, item):
        self.items.append(item)

    def removeFront(self):
        return self.items.pop()

    def removeRear(self):
        return self.items.pop(0)

    def size(self):
        return len(self.items)

# 利用Deque解决回文字符串
def palcheker(aString):
    chardeque = Deque()

    for ch in aString:
        chardeque.addFront(ch)

    stillEqual = True

    while chardeque.size() > 1 and stillEqual:
        first = chardeque.removeFront()
        last = chardeque.removeRear()
        if first != last:
            stillEqual = False

    return stillEqual

print(palcheker('lsdkjfskf'))
print(palcheker('radar'))

4
Susan
False
True

定义堆栈

# 定义一个栈类
class Stack():
    # 栈的初始化
    def __init__(self):
        self.items = []
    # 判断栈是否为空,为空返回True
    def isEmpty(self):
        return self.items ==[]
    # 向栈内压入一个元素
    def push(self, item):
        self.items.append(item)
    # 从栈内推出最后一个元素
    def pop(self):
        return self.items.pop()
    # 返回栈顶元素
    def peek(self):
        return self.items[len(self.items)-1]
    # 判断栈的大小
    def size(self):
        return len(self.items)

# 栈属性测试

s = Stack()
print(s.isEmpty())
s.push(4)
s.push('dog')
print(s.peek())
s.push(True)
print(s.isEmpty())
s.push(8.4)
print(s.pop())
print(s.pop())
print(s.size())

# 利用栈将字串的字符反转
def revstring(mystr):
    # your code here
    s = Stack()
    outputStr = ''
    for c in mystr:
        s.push(c)
    while not s.isEmpty():
        outputStr += s.pop()
    return outputStr

print(revstring('apple'))
print(revstring('x'))
print(revstring('1234567890'))

# 利用栈判断括号平衡Balanced parentheses
def parChecker(symbolString):
    s = Stack()
    balanced = True
    index = 0
    while index < len(symbolString) and balanced:
        symbol = symbolString[index]
        if symbol in '([{':
            s.push(symbol)
        else:
            if s.isEmpty():
                balanced = False
            else:
                top = s.pop()
                if not matches(top, symbol):
                    balanced = False
        index += 1

    if balanced and s.isEmpty():
        return True
    else:
        return False

def matches(open, close):
    opens = '([{'
    closers = ')]}'
    return opens.index(open) == closers.index(close)

print(parChecker('({([()])}){}'))

True
dog
False
8.4
True
2
elppa
x
0987654321
True