Python常见编程面试题

红黑树

红黑树与AVL的比较：

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: 1 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
    while a < n:
        print a,
        a, b  = b, a + b
    print
fib(1000)

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: i 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。

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 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)

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

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

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