python编程题

1、台阶问题、斐波那契

一只青蛙可以跳上一级台阶，也可以跳上两级台阶，求青蛙跳上一个n级台阶共有多少种跳法

方法一：

 def fib(n):
    a, b = 0,1
    for _ in range(n):
         a, b = b, a+b
    return b

方法二：

def fib(n):
    a, b = 0,1
    for _ in range(n):
        a, b = b, a+b
    return b

方法三：

def memo(func):
    cache = {}
    def wrap(*args):
        if args not in cache:
            cache[args] = func(*args)
        return cache[args]
    return wrap

@memo
def fib(n):
    if n < 2:
        return 1
    return fib(n-1) + fib(n-2)

2、台阶问题、斐波那契

一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。
求该青蛙跳上一个n级的台阶总共有多少种跳法。

fib = lambda n : n if n < 2 else 2*fib(n-1)

3、矩形覆盖问题

第2n个矩形的覆盖方法等于第2(n-1)加上第2(n-2)的方法。

area = lambda i :i if i < 2 else area(n-1) + area(n-2)

4、杨氏矩阵查找

在一个m行n列二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。
请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。

data = [[1,2,8,9],[2,4,9,12],[4,7,10,13],[6,8,11,15]]
def find_young(data,target):
    m = len(data) - 1
    n = len(data[0]) - 1
    r = 0
    c = n
    while c>=0 and r<=m: #从左上方开始查询
        value = data[r][c]
        if value == target:
            return True
        elif value > target:
            c -= 1
        elif value < target:
            r += 1
    return False

5、去除列表中的重复元素

list1 = ['b','c','d','b','c','a','a']

• 用集合

  list(set(data))
  

• 用字典

  l2 = {}.fromkeys(list1).keys()
    #说明：用于创建一个新字典，以序列seq中元素做字典的键
  

• 用字典并保持顺序

  l2 = list(set(list1))
  l2.sort(key=l1.index)
  

• 列表推导式

  l2 = []
  [l2.append(i) for i in list1 if not i in l2]
  

6、链表成对

1->2->3->4转换成2->1->4->3

obj = Solution()
for i in range(10):
    obj.append(i)

class ListNode:
    def __init__(self, x):
        self.value = x
        self.next = None

class Solution:
    #@param a ListNode
    #@return a ListNode
    def swapPairs(self, head):
        if head != None and head.next != None:
            next = head.next
            head.next = self.swapPairs(next.next)
            next.next = head
            return next
        return head

7、创建字典方法

工厂方法

items = [('name','earth'), ('port','80')]
dict2 = dict(items)

fromkeys()方法

dict1 = {}.fromkeys(('x', 'y'),-1)

8、合并两个有序列表

合并排序O(nlogn)

list1 = [33, 37, 38, 39]
list2 = [35, 36, 40, 43, 45, 50]

尾递归

def _recursion_merge_sort(list1, list2, tmp):
    if len(list1) == 0 or len(list2) == 0:
        tmp.extend(list1)
        tmp.extend(list2)
        return tmp
    else:
        if list1[0] <list2[0]:
            tmp.append(list1[0])
            del list1[0]
        else:
            tmp.append(list2[0])
            del list2[0]
        return _recursion_merge_sort(list1, list2, tmp)

def recursion_merge_sort(list1, list2):
    tmp = _recursion_merge_sort(list1, list2,[])
    return tmp

循环

def loop_merge_sort(list1, list2):
    result = []
    while list1 and list2:
        if list1[0] <= list2[0]:
            result.append(list1[0])
            del list1[0]
        else:
            result.append(list2[0])
            del list2[0]
    if list1 == []:
        result.extend(list2)
    if list2 == []:
        result.extend(list1)
    return result

9、交叉链表求交点

其实思想可以按照从尾开始比较两个链表，如果相交，则从尾开始必然一致，只要从尾开始比较，直至不一致的地方即为交叉点。

a = [1,2,3,7,9,1,5]
b = [4,5,7,9,1,5]
for i in range(1, min(len(a),len(b))):
    if i== 1 and (a[-1] != b[-1]):
        print('No')
        break
    else:
        if a[-i] != b[-i]:
            print('交叉节点：', a[-i+1])
            break
            
#构造链表类
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

#求出链表长度的差值，长链表的指针先想后移动lenA-lenB。
#然后两个链表一起往后走，若结点相同则第一个相交点           
def node(l1, l2):
    len1, len2 = 0, 0
    #求两链表长度
    while l1.next: 
        l1 = l1.next
        len1 += 1
    while l2.next:
        l2 = l2.next
        len2 += 1
    #如果相交
    if l1.next == l2.next:
        #长链表先走
        if len1 > len2:
            for _ in range(len1 - len2):
                l1 = l1.next
            return l1
        else:
            for _ in range(len2-len1):
                l2 = l2.next
            return l2
    else:
        return 

10、二分查找

O(logn)

def binary_search(list, item):
    low = 0
    high = len(list) - 1
    while low <= high:
        middle = int((low + high)/2)
        guess = list[middle]
        if guess > item:
            high = middle - 1
        elif guess < item:
            low = middle + 1
        else:
            return middle
    return 'Not Found'

# 排序列表
orded_list = [1,3,4,5,6,7,8,9,11]
print(binary_search(orded_list,3))
        
#补充：
l = [3,7,9,6,4,8,11,1,5]
l = sorted(l)

11、快速排序

O(nlogn)

def quick_sort(list):
    if len(list) < 2:
        return list
    else:
        middle = list[0]
        lessbeforemiddle = [x for x in list[1:] if x <= middle]
        largebefoemiddle = [x for x in list[1:] if x > middle]
        finally_list = quick_sort(lessbeforemiddle) + [middle] + quick_sort(largebefoemiddle)
        return finally_list

quick_sort(l)

12、冒泡排序

O(n^2)

def swap(l, i, j): #交换
    t = l[i]
    l[i] = l[j]
    l[j] = t
    
def bubble_sort(list):
    n = len(list)
    while n > 1:
        i = 1
        while i < n:
            if list[i-1] > list[i]:
                swap(list, i ,i-1)
            i += 1
        n -= 1

13、广度优先搜索

O(节点数+边数)

graph = {}
graph["you"] = ["alice",'bob',"calm"]
graph["alice"] = ["peggym"]
graph["bob"] = ["anuj","peggym"]
graph["peggym"] = ["anuj"]
graph["anuj"] = ["peggym"]
# print type(graph)
from collections import deque

def person_is_seller(person):
    if 'm' in person:
        return person

def search(name):
    search_queue = deque()
    search_queue += graph[name]
    searched = []
    while search_queue:
        person = search_queue.popleft()
        if person not in searched:
            if person_is_seller(person):
                print(person + 'is seller')
                searched.append(person)
            esle:
                search_queue += graph[person]
                searched.append(person)
    return False
                
search('you')

14、动态规划———找零问题

#values是硬币的面值values = [ 25, 21, 10, 5, 1]
#valuesCounts   钱币对应的种类数,5
#money  总钱数,63
#coinsUsed   对应于目前钱币总数i所使用钱币数目
def coin_change(values, valuesCounts, money, coinsUsed):
    for i in range(1, money + 1):
        minValue = i
        for num in range(valuesCounts):
            if values[num] <= i:
                count = coinsUsed[i - values[num]] + 1
                if count < minValue:
                    minValue = count
        coinsUsed[i] = minValue
        print('第%d 最少需要 %d 枚钱币' %(i, minValue))
        
 
if __name__ == '__main__':
# values = [25, 21, 10, 5, 1]
money = 63
values = [1,2,5,7,9,20]
coinsUsed = [0]*(money + 1)
length = len(values)
coin_change(values, length, money, coinsUsed)

15、二叉树节点

class Node(object):
    def __init__(self, data, left=None, right=None):
        self.data = data
        self.left = left
        self.right = right

tree = Node(1, Node(3, Node(7, 0), 6), Node(2, 5, 4)