面试题总结 - 算法

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面试题总结-算法

• 编程题
  • 1 台阶问题/斐波那契
  • 2 变态台阶问题
  • 3 矩形覆盖
  • 4 杨氏矩阵查找
  • 5 去除列表中的重复元素
  • 6 链表成对调换
  • 7 创建字典的方法
    • 1 直接创建
    • 2 工厂方法
    • 3 fromkeys()方法
  • 8 合并两个有序列表
  • 9 交叉链表求交点
  • 10 二分查找
  • 11 快排
  • 12 找零问题
  • 13 广度遍历和深度遍历二叉树
  • 17 前中后序遍历
  • 18 求最大树深
  • 19 求两棵树是否相同
  • 20 前序中序求后序
  • 21 单链表逆置
  • 22 两个字符串是否是变位词
  • 23 动态规划问题

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

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# 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

思路: http://humaoli.blog.163.com/blog/static/13346651820141125102125995/

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)

参考: http://blog.csdn.net/u013205877/article/details/76411718

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])

更多排序问题可见：数据结构与算法-排序篇-Python描述

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]))

思路: http://blog.csdn.net/wdxin1322/article/details/9501163

方法: http://www.cnblogs.com/ChenxofHit/archive/2011/03/18/1988431.html

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

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

思路: http://blog.csdn.net/feliciafay/article/details/6841115

方法: http://www.xuebuyuan.com/2066385.html?mobile=1

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

可参考：动态规划(DP)的整理-Python描述

http://fm126.top/article/30