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
