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 printfib(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 = Noneclass 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 tmp9 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 = Nonedef 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 Falseif __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]+greaterif __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 = righttree = 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 curdef 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 = nextlink = 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 preroot = rev(link)while root: print root.data root = root.next
