Python面试--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。

Python

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

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

第二种记忆方法

Python

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)

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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

Python

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

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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。

Python

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

1	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。

Python

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

1	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。

5 去除列表中的重复元素

用集合

Python

list(set(l))

1	list(set(l))

用字典

Python

l1 = ['b','c','d','b','c','a','a'] l2 = {}.fromkeys(l1).keys() print l2

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l1 = ['b','c','d','b','c','a','a'] l2 = {}.fromkeys(l1).keys() print l2

用字典并保持顺序

Python

l1 = ['b','c','d','b','c','a','a'] l2 = list(set(l1)) l2.sort(key=l1.index) print l2

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l1 = ['b','c','d','b','c','a','a'] l2 = list(set(l1)) l2.sort(key=l1.index) print l2

列表推导式

Python

l1 = ['b','c','d','b','c','a','a'] l2 = [] [l2.append(i) for i in l1 if not i in l2]

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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.

Python

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

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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

Python

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

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

2 工厂方法

Python

items=[('name','earth'),('port','80')] dict2=dict(items) dict1=dict((['name','earth'],['port','80']))

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items=[('name','earth'),('port','80')] dict2=dict(items) dict1=dict((['name','earth'],['port','80']))

3 fromkeys()方法

Python

dict1={}.fromkeys(('x','y'),-1) dict={'x':-1,'y':-1} dict2={}.fromkeys(('x','y')) dict2={'x':None, 'y':None}

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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

Python

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, [])

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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

Python

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

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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.

Python

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

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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

Python

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)

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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

Python

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))

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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

Python

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)

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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

Python

## 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)

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## 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

Python

def maxDepth(root): if not root: return 0 return max(maxDepth(root.left), maxDepth(root.right)) + 1

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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

Python

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

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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.youkuaiyun.com/hinyunsin/article/details/6315502

Python

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

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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

Python

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

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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