2. Python面试编程题汇总
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编程题
1 台阶问题/斐波纳挈
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]
面试官提到的,先排序然后删除.
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
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
9 交叉链表求交点
去哪儿的面试,没做出来.
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
10 二分查找
def binarySearch(l, t):
low, high = 0, len(l) - 1
while low < high:
print low, high
mid = (low + high) / 2
if l[mid] > t:
high = mid
elif l[mid] < t:
low = mid + 1
else:
return mid
return low if l[low] == t else False
if __name__ == ‘__main__‘:
l = [1, 4, 12, 45, 66, 99, 120, 444]
print binarySearch(l, 12)
print binarySearch(l, 1)
print binarySearch(l, 13)
print binarySearch(l, 444)
11 快排
def qsort(seq):
if seq==[]:
return []
else:
pivot=seq[0]
lesser=qsort([x for x in seq[1:] if x<pivot])
greater=qsort([x for x in seq[1:] if x>=pivot])
return lesser+[pivot]+greater
if __name__==‘__main__‘:
seq=[5,6,78,9,0,-1,2,3,-65,12]
print(qsort(seq))
12 找零问题
def coinChange(values, money, coinsUsed):
#values T[1:n]数组
#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
for cents in range(1, money+1):
minCoins = cents #从第一个开始到money的所有情况初始
for value in values:
if value <= cents:
temp = coinsUsed[cents - value] + 1
if temp < minCoins:
minCoins = temp
coinsUsed[cents] = minCoins
print(‘面值为：{0} 的最小硬币数目为：{1} ‘.format(cents, coinsUsed[cents]) )
if __name__ == ‘__main__‘:
values = [ 25, 21, 10, 5, 1]
money = 63
coinsUsed = {i:0 for i in range(money+1)}
coinChange(values, money, coinsUsed)
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):
stack = [root]
while stack:
current = stack.pop(0)
print current.data
if current.left:
stack.append(current.left)
if current.right:
stack.append(current.right)
## 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
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
推荐: http://blog.csdn.net/hinyunsin/article/details/6315502
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‘))
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