编程题 复习

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