一刷时间：6月8日--6月9日
用栈实现队列
python
class MyQueue(object):
def __init__(self):
self.stack_in=[]
self.stack_out=[]
def push(self, x):
self.stack_in.append(x)
def pop(self):
if self.stack_out:
return self.stack_out.pop()
else:
while self.stack_in:
self.stack_out.append(self.stack_in.pop())
return self.stack_out.pop()
def peek(self):
x=self.pop()
self.stack_out.append(x)
return x
def empty(self):
if self.stack_in or self.stack_out:
return False
else:
return True用队列实现栈
python
class MyStack(object):
def __init__(self):
self.deque=deque()
def push(self, x):
"""
:type x: int
:rtype: None
"""
self.deque.append(x)
def pop(self):
"""
:rtype: int
"""
return self.deque.pop()
def top(self):
"""
:rtype: int
"""
x=self.deque.pop()
self.deque.append(x)
return x
def empty(self):
"""
:rtype: bool
"""
if self.deque:
return False
else:
return True有效的括号
python
alpha=[]
for x in s:
if x=='(' or x=='[' or x=='{':
alpha.append(x)
else:
if x==')':
if len(alpha)==0 or alpha.pop()!='(':
return False
elif x=='}':
if len(alpha)==0 or alpha.pop()!='{':
return False
elif x==']':
if len(alpha)==0 or alpha.pop()!='[':
return False
if len(alpha)==0:
return True
else:
return False删除字符串中所有相邻重复项
python
re=[]
for x in s:
if len(re)==0:
re.append(x)
else:
if re[-1]==x:
re.pop()
else:
re.append(x)
return "".join(re)逆波兰表达式求值
使用python评测通过，主要区别就是python和python3在处理除法运算的不同处理
python
nums=[]
for x in tokens:
if x=='+' or x=='-' or x=='*' or x=='/':
x2=int(nums.pop())
x1=int(nums.pop())
if x=='+':
nums.append(x1+x2)
elif x=='-':
nums.append(x1-x2)
elif x=='*':
nums.append(x1*x2)
elif x=='/':
nums.append(int(float(x1)/x2))
else:
nums.append(int(x))
return nums[0]使用python3运行通过
python
nums=[]
for x in tokens:
if x=='+' or x=='-' or x=='*' or x=='/':
x2=int(nums.pop())
x1=int(nums.pop())
if x=='+':
nums.append(x1+x2)
elif x=='-':
nums.append(x1-x2)
elif x=='*':
nums.append(x1*x2)
elif x=='/':
nums.append(int(x1/x2)) #向零取整
else:
nums.append(int(x))
return nums[0]滑动窗口最大值
基本思想是维持一个最大窗口，判断被弹出和被推入的是否是最大值，进行相应的处理。
本质上应该是n*k的复杂度，超过时间限制了。。。
python
array=[]
for i in range(len(nums)-k+1):
if i-1>=0 and nums[i-1]<array[-1]:
if i+k-1<len(nums) and nums[i+k-1]<=array[-1]:
array.append(array[-1])
elif i+k-1<len(nums) and nums[i+k-1]>array[-1]:
array.append(nums[i+k-1])
else:
array.append(max(nums[i:i+k]))
elif i-1>=0 and nums[i-1]==array[-1]:
if i+k-1<len(nums) and nums[i+k-1]<=nums[i-1]:
array.append(max(nums[i:i+k]))
elif i+k-1<len(nums) and nums[i+k-1]>nums[i-1]:
array.append(nums[i+k-1])
else:
array.append(max(nums[i:i+k]))
return array自定义一个队列，保留可能是最大值的数
python
class MyDeque():
def __init__(self):
self.deque=deque()
def pop(self,value):
if self.deque and self.deque[0]==value:
self.deque.popleft()
def push(self,value):
while self.deque and value>self.deque[-1]:
self.deque.pop()
self.deque.append(value)
def getmax(self):
return self.deque[0]
class Solution(object):
def maxSlidingWindow(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
mydeque=MyDeque()
result=[]
for i in range(k):
mydeque.push(nums[i])
result.append(mydeque.getmax())
for i in range(k,len(nums)):
mydeque.pop(nums[i-k])
mydeque.push(nums[i])
result.append(mydeque.getmax())
return result
前k个高频元素
使用字典统计每个字符出现频率，并使用最小堆维持前k个最大的频率
python
map_={}
for i in nums:
map_[i]=map_.get(i,0)+1
pre_que=[]
for idx,freq in map_.items():
heapq.heappush(pre_que,(freq,idx))
if len(pre_que)>k:
heapq.heappop(pre_que)
reslut=[]
for i in range(k):
reslut.append(heapq.heappop(pre_que)[1])
return reslut