编程题

1 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。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 fib(n - 1) + fib(n - 2)

第二种记忆方法

def memo(func):

cache = {}

def wrap(*args):

if args not in cache:

cache[args] = func(*args)

return cache[args]

return wrap

@memo

def fib(i):

if i < 2:

return 1

return fib(i-1) + fib(i-2)

第三种方法

def fib(n):

a, b = 0, 1

for _ in xrange(n):

a, b = b, a + b

return b

2 变态台阶问题

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

我们可以用2*1的小矩形横着或者竖着去覆盖更大的矩形。请问用n个2*1的小矩形无重叠地覆盖一个2*n的大矩形，总共有多少种方法？

第2*n个矩形的覆盖方法等于第2*(n-1)加上第2*(n-2)的方法。

f = lambda n: 1 if n < 2 else f(n - 1) + f(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。

使用Step-wise线性搜索。

def get_value(l, r, c):

return l[r][c]

def find(l, x):

m = len(l) - 1

n = len(l[0]) - 1

r = 0

c = n

while c >= 0 and r <= m:

value = get_value(l, r, c)

if value == x:

return True

elif value > x:

c = c - 1

elif value < x:

r = r + 1

return False

5 去除列表中的重复元素

用集合

list(set(l))

用字典

l1 = ['b','c','d','b','c','a','a']

l2 = {}.fromkeys(l1).keys()

print l2

用字典并保持顺序

l1 = ['b','c','d','b','c','a','a']

l2 = list(set(l1))

l2.sort(key=l1.index)

print l2

列表推导式

l1 = ['b','c','d','b','c','a','a']

l2 = []

[l2.append(i) for i in l1 if not i in l2]

sorted排序并且用列表推导式.

l = ['b','c','d','b','c','a','a'] [single.append(i) for i in sorted(l) if i not in single] print single

6 链表成对调换

1->2->3->4转换成2->1->4->3.

class ListNode:

def __init__(self, x):

self.val = 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 创建字典的方法

1 直接创建

dict = {'name':'earth', 'port':'80'}

2 工厂方法

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

dict2=dict(items)

dict1=dict((['name','earth'],['port','80']))

3 fromkeys()方法

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

dict={'x':-1,'y':-1}

dict2={}.fromkeys(('x','y'))

dict2={'x':None, 'y':None}

8 合并两个有序列表

知乎远程面试要求编程

尾递归

def _recursion_merge_sort2(l1, l2, tmp):

if len(l1) == 0 or len(l2) == 0:

tmp.extend(l1)

tmp.extend(l2)

return tmp

else:

if l1[0] < l2[0]:

tmp.append(l1[0])

del l1[0]

else:

tmp.append(l2[0])

del l2[0]

return _recursion_merge_sort2(l1, l2, tmp)

def recursion_merge_sort2(l1, l2):

return _recursion_merge_sort2(l1, l2, [])

循环算法

思路：

定义一个新的空列表

比较两个列表的首个元素

小的就插入到新列表里

把已经插入新列表的元素从旧列表删除

直到两个旧列表有一个为空

再把旧列表加到新列表后面

def loop_merge_sort(l1, l2):

tmp = []

while len(l1) > 0 and len(l2) > 0:

if l1[0] < l2[0]:

tmp.append(l1[0])

del l1[0]

else:

tmp.append(l2[0])

del l2[0]

tmp.extend(l1)

tmp.extend(l2)

return tmp

pop弹出

a = [1,2,3,7]

b = [3,4,5]

def merge_sortedlist(a,b):

c = []

while a and b:

if a[0] >= b[0]:

c.append(b.pop(0))

else:

c.append(a.pop(0))

while a:

c.append(a.pop(0))

while b:

c.append(b.pop(0))

return c

print merge_sortedlist(a,b)

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，如图所示

# 使用a,b两个list来模拟链表，可以看出交叉点是 7这个节点

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

else:

pass

另外一种比较正规的方法，构造链表类

class ListNode:

def __init__(self, x):

self.val = x

self.next = None

def node(l1, l2):

length1, lenth2 = 0, 0

# 求两个链表长度

while l1.next:

l1 = l1.next

length1 += 1

while l2.next:

l2 = l2.next

length2 += 1

# 长的链表先走

if length1 > lenth2:

for _ in range(length1 - length2):

l1 = l1.next

else:

for _ in range(length2 - length1):

l2 = l2.next

while l1 and l2:

if l1.next == l2.next:

return l1.next

else:

l1 = l1.next

l2 = l2.next

修改了一下:

#coding:utf-8

class ListNode:

def __init__(self, x):

self.val = x

self.next = None

def node(l1, l2):

length1, length2 = 0, 0

# 求两个链表长度

while l1.next:

l1 = l1.next#尾节点

length1 += 1

while l2.next:

l2 = l2.next#尾节点

length2 += 1

#如果相交

if l1.next == l2.next:

# 长的链表先走

if length1 > length2:

for _ in range(length1 - length2):

l1 = l1.next

return l1#返回交点

else:

for _ in range(length2 - length1):

l2 = l2.next

return l2#返回交点

# 如果不相交

else:

return

10 二分查找

#coding:utf-8

def binary_search(list, item):

low = 0

high = len(list) - 1

while low <= high:

mid = (high - low) / 2 + low # 避免(high + low) / 2溢出

guess = list[mid]

if guess > item:

high = mid - 1

elif guess < item:

low = mid + 1

else:

return mid

return None

mylist = [1,3,5,7,9]

print binary_search(mylist, 3)

11 快排

#coding:utf-8

def quicksort(list):

if len(list)<2:

return list

else:

midpivot = list[0]

lessbeforemidpivot = [i for i in list[1:] if i<=midpivot]

biggerafterpivot = [i for i in list[1:] if i > midpivot]

finallylist = quicksort(lessbeforemidpivot)+[midpivot]+quicksort(biggerafterpivot)

return finallylist

print quicksort([2,4,6,7,1,2,5])

12 找零问题

#coding:utf-8

#values是硬币的面值values = [ 25, 21, 10, 5, 1]

#valuesCounts 钱币对应的种类数

#money 找出来的总钱数

#coinsUsed 对应于目前钱币总数i所使用的硬币数目

def coinChange(values,valuesCounts,money,coinsUsed):

#遍历出从1到money所有的钱数可能

for cents in range(1,money+1):

minCoins = cents

#把所有的硬币面值遍历出来和钱数做对比

for kind in range(0,valuesCounts):

if (values[kind] <= cents):

temp = coinsUsed[cents - values[kind]] +1

if (temp < minCoins):

minCoins = temp

coinsUsed[cents] = minCoins

print ('面值:{0}的最少硬币使用数为:{1}'.format(cents, coinsUsed[cents]))

13 广度遍历和深度遍历二叉树

给定一个数组，构建二叉树，并且按层次打印这个二叉树

14 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, Node(0)), Node(6)), Node(2, Node(5), Node(4)))

15 层次遍历

def lookup(root):

row = [root]

while row:

print(row)

row = [kid for item in row for kid in (item.left, item.right) if kid]

16 深度遍历

def deep(root):

if not root:

return

print root.data

deep(root.left)

deep(root.right)

if __name__ == '__main__':

lookup(tree)

deep(tree)

17 前中后序遍历

深度遍历改变顺序就OK了

#coding:utf-8

#二叉树的遍历

#简单的二叉树节点类

class Node(object):

def __init__(self,value,left,right):

self.value = value

self.left = left

self.right = right

#中序遍历:遍历左子树,访问当前节点,遍历右子树

def mid_travelsal(root):

if root.left is not None:

mid_travelsal(root.left)

#访问当前节点

print(root.value)

if root.right is not None:

mid_travelsal(root.right)

#前序遍历:访问当前节点,遍历左子树,遍历右子树

def pre_travelsal(root):

print (root.value)

if root.left is not None:

pre_travelsal(root.left)

if root.right is not None:

pre_travelsal(root.right)

#后续遍历:遍历左子树,遍历右子树,访问当前节点

def post_trvelsal(root):

if root.left is not None:

post_trvelsal(root.left)

if root.right is not None:

post_trvelsal(root.right)

print (root.value)

18 求最大树深

def maxDepth(root):

if not root:

return 0

return max(maxDepth(root.left), maxDepth(root.right)) + 1

19 求两棵树是否相同

def isSameTree(p, q):

if p == None and q == None:

return True

elif p and q :

return p.val == q.val and isSameTree(p.left,q.left) and isSameTree(p.right,q.right)

else :

return False

20 前序中序求后序

def rebuild(pre, center):

if not pre:

return

cur = Node(pre[0])

index = center.index(pre[0])

cur.left = rebuild(pre[1:index + 1], center[:index])

cur.right = rebuild(pre[index + 1:], center[index + 1:])

return cur

def deep(root):

if not root:

return

deep(root.left)

deep(root.right)

print root.data

21 单链表逆置

class Node(object):

def __init__(self, data=None, next=None):

self.data = data

self.next = next

link = Node(1, Node(2, Node(3, Node(4, Node(5, Node(6, Node(7, Node(8, Node(9)))))))))

def rev(link):

pre = link

cur = link.next

pre.next = None

while cur:

tmp = cur.next

cur.next = pre

pre = cur

cur = tmp

return pre

root = rev(link)

while root:

print root.data

root = root.next

22 两个字符串是否是变位词

class Anagram:

"""

@:param s1: The first string

@:param s2: The second string

@:return true or false

"""

def Solution1(s1,s2):

alist = list(s2)

pos1 = 0

stillOK = True

while pos1 < len(s1) and stillOK:

pos2 = 0

found = False

while pos2 < len(alist) and not found:

if s1[pos1] == alist[pos2]:

found = True

else:

pos2 = pos2 + 1

if found:

alist[pos2] = None

else:

stillOK = False

pos1 = pos1 + 1

return stillOK

print(Solution1('abcd','dcba'))

def Solution2(s1,s2):

alist1 = list(s1)

alist2 = list(s2)

alist1.sort()

alist2.sort()

pos = 0

matches = True

while pos < len(s1) and matches:

if alist1[pos] == alist2[pos]:

pos = pos + 1

else:

matches = False

return matches

print(Solution2('abcde','edcbg'))

def Solution3(s1,s2):

c1 = [0]*26

c2 = [0]*26

for i in range(len(s1)):

pos = ord(s1[i])-ord('a')

c1[pos] = c1[pos] + 1

for i in range(len(s2)):

pos = ord(s2[i])-ord('a')

c2[pos] = c2[pos] + 1

j = 0

stillOK = True

while j<26 and stillOK:

if c1[j] == c2[j]:

j = j + 1

else:

stillOK = False

return stillOK

print(Solution3('apple','pleap'))

23 动态规划问题