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(效率更高,相同的传参,直接得到结果,LRU),性能极好,用这个方法,让你的笔试脱颖而出。
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)
def fib(n):
a, b = 0, 1
while a < n:
print a,
a, b = b, a + b
print
fib(1000)
2.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
用字典并保持排序:
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]
还有一题经常考,字段根据key值排序
d = {'lilee':25, 'wangyan':21, 'liqun':32, 'age':19}
sorted(d.items(), key=lambda item:item[0])
字段根据value值排序
d = {'lilee':25, 'wangyan':21, 'liqun':32, 'age':19}
sorted(d.items(), key=lambda item:item[1])
3.合并两个有序列表
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, [])
4.二分查找
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)
5.快速排序
def qsort(seq):
if seq==[]:
return []
else:
pivot=seq[0]
lesser=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))
6.广度遍历和深度遍历二叉树
## 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)
求最大树深
def maxDepth(root):
if not root:
return 0
return max(maxDepth(root.left), maxDepth(root.right)) + 1
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
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