一刷时间：6月8日–6月9日

用栈实现队列

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 xdef empty(self):if self.stack_in or self.stack_out:return Falseelse:return True

用队列实现栈

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 xdef empty(self):""":rtype: bool"""if self.deque:return Falseelse:return True

有效的括号

 		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 Falseelif x=='}':if len(alpha)==0 or alpha.pop()!='{':return Falseelif  x==']':if len(alpha)==0 or alpha.pop()!='[':return Falseif len(alpha)==0:return Trueelse:return False

删除字符串中所有相邻重复项

 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

 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

		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。。。

 		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

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

  		map_={}for i in nums:map_[i]=map_.get(i,0)+1pre_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