力扣刷题

刷题学习：

160链表相交节点查找 python实现

def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
a,b = 0,0
p,q = headA,headB
while headA and headB:
headA,headB = headA.next,headB.next
a,b = a+1,b+1
while headA:
headA,p = headA.next,p.next
while headB:
headB,q = headB.next,q.next
while p:
if p==q:
return p
p,q = p.next,q.next

return
'''
if not headA or not headB:#空链表检测，无论谁空直接返回
return
p,q = headA,headB
while p!=q:#并行比较链表节点value，复杂度O(N)
p = p.next if p else headB
q = q.next if q else headA
return p
'''

241 class Solution:
def diffWaysToCompute(self, input: str) -> List[int]:

#241 分治
res = []
ops = {'+':lambda x,y:x+y, '-':lambda x,y:x-y, '*':lambda x,y:x*y}
for indx in range(1,len(input)-1):
if input[indx] in ops.keys():
for left in self.diffWaysToCompute(input[:indx]):
for right in self.diffWaysToCompute(input[indx+1:]):
res.append(ops[input[indx]](left,right))
if not res:
res.append(int(input))

return res

242 字符串判断是否异位相同class Solution:
def isAnagram(self, s: str, t: str) -> bool:
if len(s)!=len(t):
return False
s_=list(s)
t_=list(t)
s_res=dict.fromkeys(s_,0)
t_res=dict.fromkeys(t_,0)
for i in s_:
if i in s_res.keys():
s_res[i]+=1
for j in t_:
if j in t_res.keys():
t_res[j]+=1
if len(s_res.items()&t_res.items())==len(s_res):
return True
else:
return False

232 栈队列

class MyQueue(object):

    def __init__(self):
     
        self.input = []
        self.output = []
        

    def push(self, x):
   
        self.input.append(x)

    def pop(self):
    
        if  self.output:
            return self.output.pop()
        else:
            while self.input:
                self.output.append(self.input.pop())
            return self.output.pop()
        
    def peek(self):
      
        if  self.output:
            return self.output[-1]
        else:
            while self.input:
                self.output.append(self.input.pop())
            return self.output[-1]
    def empty(self):
        
        return len(self.input) == 0 and len(self.output) == 0

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