mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-08 16:54:50 +08:00
程序员Carl
@ -162,18 +162,44 @@ class Solution {
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Python：
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```python3
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# 方法一，仅使用栈，更省空间
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class Solution:
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def isValid(self, s: str) -> bool:
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stack = [] # 保存还未匹配的左括号
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mapping = {")": "(", "]": "[", "}": "{"}
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for i in s:
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if i in "([{": # 当前是左括号，则入栈
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stack.append(i)
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elif stack and stack[-1] == mapping[i]: # 当前是配对的右括号则出栈
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stack.pop()
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else: # 不是匹配的右括号或者没有左括号与之匹配，则返回false
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stack = []
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for item in s:
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if item == '(':
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stack.append(')')
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elif item == '[':
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stack.append(']')
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elif item == '{':
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stack.append('}')
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elif not stack or stack[-1] != item:
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return False
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return stack == [] # 最后必须正好把左括号匹配完
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else:
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stack.pop()
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return True if not stack else False
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```
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```python3
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# 方法二，使用字典
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class Solution:
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def isValid(self, s: str) -> bool:
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stack = []
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mapping = {
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'(': ')',
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'[': ']',
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'{': '}'
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}
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for item in s:
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if item in mapping.keys():
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stack.append(mapping[item])
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elif not stack or stack[-1] != item:
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return False
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else:
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stack.pop()
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return True if not stack else False
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```
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Go：
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@ -160,21 +160,29 @@ class Solution {
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Python：
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def swapPairs(self, head: ListNode) -> ListNode:
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dummy = ListNode(0) #设置一个虚拟头结点
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dummy.next = head
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cur = dummy
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while cur.next and cur.next.next:
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tmp = cur.next #记录临时节点
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tmp1 = cur.next.next.next #记录临时节点
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res = ListNode(next=head)
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pre = res
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# 必须有pre的下一个和下下个才能交换，否则说明已经交换结束了
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while pre.next and pre.next.next:
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cur = pre.next
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post = pre.next.next
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cur.next = cur.next.next #步骤一
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cur.next.next = tmp #步骤二
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cur.next.next.next = tmp1 #步骤三
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cur = cur.next.next #cur移动两位，准备下一轮交换
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return dummy.next
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# pre，cur，post对应最左，中间的，最右边的节点
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cur.next = post.next
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post.next = cur
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pre.next = post
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pre = pre.next.next
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return res.next
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```
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Go：
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@ -223,17 +223,19 @@ var evalRPN = function(tokens) {
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python3
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```python
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def evalRPN(tokens) -> int:
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stack = list()
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for i in range(len(tokens)):
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if tokens[i] not in ["+", "-", "*", "/"]:
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stack.append(tokens[i])
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else:
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tmp1 = stack.pop()
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tmp2 = stack.pop()
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res = eval(tmp2+tokens[i]+tmp1)
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stack.append(str(int(res)))
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return stack[-1]
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class Solution:
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def evalRPN(self, tokens: List[str]) -> int:
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stack = []
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for item in tokens:
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if item not in {"+", "-", "*", "/"}:
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stack.append(item)
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else:
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first_num, second_num = stack.pop(), stack.pop()
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stack.append(
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int(eval(f'{second_num} {item} {first_num}')) # 第一个出来的在运算符后面
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)
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return int(stack.pop()) # 如果一开始只有一个数，那么会是字符串形式的
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```
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@ -245,13 +245,15 @@ Python：
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def removeElements(self, head: ListNode, val: int) -> ListNode:
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dummy_head = ListNode(next=head) #添加一个虚拟节点
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dummy_head = ListNode(next=head)
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cur = dummy_head
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while(cur.next!=None):
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if(cur.next.val == val):
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cur.next = cur.next.next #删除cur.next节点
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while cur.next:
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if cur.next.val == val:
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cur.next = cur.next.next # 删除下一个节点
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else:
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cur = cur.next
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return dummy_head.next
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@ -294,53 +294,66 @@ Python：
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```python
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from collections import deque
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class MyStack:
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def __init__(self):
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"""
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Initialize your data structure here.
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Python普通的Queue或SimpleQueue没有类似于peek的功能
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也无法用索引访问，在实现top的时候较为困难。
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用list可以，但是在使用pop(0)的时候时间复杂度为O(n)
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因此这里使用双向队列，我们保证只执行popleft()和append()，因为deque可以用索引访问，可以实现和peek相似的功能
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in - 存所有数据
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out - 仅在pop的时候会用到
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"""
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#使用两个队列来实现
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self.que1 = deque()
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self.que2 = deque()
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self.queue_in = deque()
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self.queue_out = deque()
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def push(self, x: int) -> None:
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"""
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Push element x onto stack.
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直接append即可
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"""
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self.que1.append(x)
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self.queue_in.append(x)
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def pop(self) -> int:
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"""
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Removes the element on top of the stack and returns that element.
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1. 首先确认不空
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2. 因为队列的特殊性，FIFO，所以我们只有在pop()的时候才会使用queue_out
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3. 先把queue_in中的所有元素（除了最后一个），依次出列放进queue_out
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4. 交换in和out，此时out里只有一个元素
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5. 把out中的pop出来，即是原队列的最后一个
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tip：这不能像栈实现队列一样，因为另一个queue也是FIFO，如果执行pop()它不能像
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stack一样从另一个pop()，所以干脆in只用来存数据，pop()的时候两个进行交换
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"""
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size = len(self.que1)
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size -= 1#这里先减一是为了保证最后面的元素
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while size > 0:
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size -= 1
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self.que2.append(self.que1.popleft())
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if self.empty():
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return None
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result = self.que1.popleft()
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self.que1, self.que2= self.que2, self.que1#将que2和que1交换 que1经过之前的操作应该是空了
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#一定注意不能直接使用que1 = que2 这样que2的改变会影响que1 可以用浅拷贝
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return result
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for i in range(len(self.queue_in) - 1):
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self.queue_out.append(self.queue_in.popleft())
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self.queue_in, self.queue_out = self.queue_out, self.queue_in # 交换in和out，这也是为啥in只用来存
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return self.queue_out.popleft()
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def top(self) -> int:
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"""
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Get the top element.
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1. 首先确认不空
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2. 我们仅有in会存放数据，所以返回第一个即可
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"""
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return self.que1[-1]
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if self.empty():
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return None
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return self.queue_in[-1]
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def empty(self) -> bool:
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"""
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Returns whether the stack is empty.
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因为只有in存了数据，只要判断in是不是有数即可
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"""
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#print(self.que1)
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if len(self.que1) == 0:
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return True
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else:
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return False
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return len(self.queue_in) == 0
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```
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@ -281,48 +281,60 @@ class MyQueue {
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Python：
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```python
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# 使用两个栈实现先进先出的队列
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class MyQueue:
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def __init__(self):
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"""
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Initialize your data structure here.
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in主要负责push，out主要负责pop
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"""
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self.stack1 = list()
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self.stack2 = list()
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self.stack_in = []
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self.stack_out = []
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def push(self, x: int) -> None:
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"""
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Push element x to the back of queue.
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有新元素进来，就往in里面push
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"""
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# self.stack1用于接受元素
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self.stack1.append(x)
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self.stack_in.append(x)
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def pop(self) -> int:
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"""
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Removes the element from in front of queue and returns that element.
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1. 检查如果out里面元素，则直接pop
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2. 如果out没有元素，就把in里面的元素（除了第一个）依次pop后装进out里面
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3. 直接把in剩下的元素pop出来，就是queue头部的
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"""
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# self.stack2用于弹出元素，如果self.stack2为[],则将self.stack1中元素全部弹出给self.stack2
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if self.stack2 == []:
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while self.stack1:
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tmp = self.stack1.pop()
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self.stack2.append(tmp)
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return self.stack2.pop()
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if self.empty:
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return None
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if self.stack_out:
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return self.stack_out.pop()
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else:
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for i in range(1, len(self.stack_in)):
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self.stack_out.append(self.stack_in.pop())
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return self.stack_in.pop()
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def peek(self) -> int:
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"""
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Get the front element.
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1. 查out有没有元素，有就把最上面的返回
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2. 如果out没有元素，就把in最下面的返回
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"""
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if self.stack2 == []:
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while self.stack1:
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tmp = self.stack1.pop()
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self.stack2.append(tmp)
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return self.stack2[-1]
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if self.empty:
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return None
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if self.stack_out:
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return self.stack_out[-1]
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else:
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return self.stack_in[0]
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def empty(self) -> bool:
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"""
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Returns whether the queue is empty.
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只要in或者out有元素，说明队列不为空
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"""
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return self.stack1 == [] and self.stack2 == []
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return not (self.stack_in or self.stack_out)
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```
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@ -162,21 +162,14 @@ class Solution:
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Do not return anything, modify s in-place instead.
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"""
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left, right = 0, len(s) - 1
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while(left < right):
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# 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低
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# 推荐该写法，更加通俗易懂
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while left < right:
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s[left], s[right] = s[right], s[left]
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left += 1
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right -= 1
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# 下面的写法更加简洁，但是都是同样的算法
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# class Solution:
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# def reverseString(self, s: List[str]) -> None:
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# """
|
||||
# Do not return anything, modify s in-place instead.
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# """
|
||||
# 不需要判别是偶数个还是奇数个序列，因为奇数个的时候，中间那个不需要交换就可
|
||||
# for i in range(len(s)//2):
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# s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i]
|
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# return s
|
||||
|
||||
```
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Go：
|
||||
|
||||
@ -155,34 +155,27 @@ class Solution {
|
||||
|
||||
Python：
|
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```python
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||||
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class Solution(object):
|
||||
def reverseStr(self, s, k):
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class Solution:
|
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def reverseStr(self, s: str, k: int) -> str:
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||||
"""
|
||||
:type s: str
|
||||
:type k: int
|
||||
:rtype: str
|
||||
1. 使用range(start, end, step)来确定需要调换的初始位置
|
||||
2. 对于字符串s = 'abc'，如果使用s[0:999] ===> 'abc'。字符串末尾如果超过最大长度，则会返回至字符串最后一个值，这个特性可以避免一些边界条件的处理。
|
||||
3. 用切片整体替换，而不是一个个替换.
|
||||
"""
|
||||
from functools import reduce
|
||||
# turn s into a list
|
||||
s = list(s)
|
||||
|
||||
# another way to simply use a[::-1], but i feel this is easier to understand
|
||||
def reverse(s):
|
||||
left, right = 0, len(s) - 1
|
||||
def reverse_substring(text):
|
||||
left, right = 0, len(text) - 1
|
||||
while left < right:
|
||||
s[left], s[right] = s[right], s[left]
|
||||
text[left], text[right] = text[right], text[left]
|
||||
left += 1
|
||||
right -= 1
|
||||
return s
|
||||
return text
|
||||
|
||||
# make sure we reverse each 2k elements
|
||||
for i in range(0, len(s), 2*k):
|
||||
s[i:(i+k)] = reverse(s[i:(i+k)])
|
||||
|
||||
# combine list into str.
|
||||
return reduce(lambda a, b: a+b, s)
|
||||
res = list(s)
|
||||
|
||||
for cur in range(0, len(s), 2 * k):
|
||||
res[cur: cur + k] = reverse_substring(res[cur: cur + k])
|
||||
|
||||
return ''.join(res)
|
||||
```
|
||||
|
||||
|
||||
|
||||
@ -197,15 +197,38 @@ class Solution {
|
||||
|
||||
Python：
|
||||
```python3
|
||||
# 方法一，使用栈，推荐！
|
||||
class Solution:
|
||||
def removeDuplicates(self, s: str) -> str:
|
||||
t = list()
|
||||
for i in s:
|
||||
if t and t[-1] == i:
|
||||
t.pop(-1)
|
||||
res = list()
|
||||
for item in s:
|
||||
if res and res[-1] == item:
|
||||
res.pop()
|
||||
else:
|
||||
t.append(i)
|
||||
return "".join(t) # 字符串拼接
|
||||
res.append(item)
|
||||
return "".join(res) # 字符串拼接
|
||||
```
|
||||
|
||||
```python3
|
||||
# 方法二，使用双指针模拟栈，如果不让用栈可以作为备选方法。
|
||||
class Solution:
|
||||
def removeDuplicates(self, s: str) -> str:
|
||||
res = list(s)
|
||||
slow = fast = 0
|
||||
length = len(res)
|
||||
|
||||
while fast < length:
|
||||
# 如果一样直接换，不一样会把后面的填在slow的位置
|
||||
res[slow] = res[fast]
|
||||
|
||||
# 如果发现和前一个一样，就退一格指针
|
||||
if slow > 0 and res[slow] == res[slow - 1]:
|
||||
slow -= 1
|
||||
else:
|
||||
slow += 1
|
||||
fast += 1
|
||||
|
||||
return ''.join(res[0: slow])
|
||||
```
|
||||
|
||||
Go：
|
||||
|
||||
@ -202,45 +202,27 @@ func replaceSpace(s string) string {
|
||||
|
||||
python：
|
||||
```python
|
||||
class Solution(object):
|
||||
def replaceSpace(self, s):
|
||||
"""
|
||||
:type s: str
|
||||
:rtype: str
|
||||
"""
|
||||
list_s = list(s)
|
||||
|
||||
# 记录原本字符串的长度
|
||||
original_end = len(s)
|
||||
|
||||
# 将空格改成%20 使得字符串总长增长 2n，n为原本空格数量。
|
||||
# 所以记录空格数量就可以得到目标字符串的长度
|
||||
n_space = 0
|
||||
for ss in s:
|
||||
if ss == ' ':
|
||||
n_space += 1
|
||||
|
||||
list_s += ['0'] * 2 * n_space
|
||||
|
||||
# 设置左右指针位置
|
||||
left, right = original_end - 1, len(list_s) - 1
|
||||
|
||||
# 循环直至左指针越界
|
||||
while left >= 0:
|
||||
if list_s[left] == ' ':
|
||||
list_s[right] = '0'
|
||||
list_s[right - 1] = '2'
|
||||
list_s[right - 2] = '%'
|
||||
right -= 3
|
||||
else:
|
||||
list_s[right] = list_s[left]
|
||||
right -= 1
|
||||
|
||||
left -= 1
|
||||
class Solution:
|
||||
def replaceSpace(self, s: str) -> str:
|
||||
counter = s.count(' ')
|
||||
|
||||
# 将list变回str，输出
|
||||
s = ''.join(list_s)
|
||||
return s
|
||||
res = list(s)
|
||||
# 每碰到一个空格就多拓展两个格子，1 + 2 = 3个位置存’%20‘
|
||||
res.extend([' '] * counter * 2)
|
||||
|
||||
# 原始字符串的末尾，拓展后的末尾
|
||||
left, right = len(s) - 1, len(res) - 1
|
||||
|
||||
while left >= 0:
|
||||
if res[left] != ' ':
|
||||
res[right] = res[left]
|
||||
right -= 1
|
||||
else:
|
||||
# [right - 2, right), 左闭右开
|
||||
res[right - 2: right + 1] = '%20'
|
||||
right -= 3
|
||||
left -= 1
|
||||
return ''.join(res)
|
||||
|
||||
```
|
||||
|
||||
|
||||
@ -125,24 +125,45 @@ python:
|
||||
class Solution:
|
||||
def reverseLeftWords(self, s: str, n: int) -> str:
|
||||
return s[n:] + s[0:n]
|
||||
|
||||
```
|
||||
```python
|
||||
# 方法二：也可以使用上文描述的方法，有些面试中不允许使用切片，那就使用上文作者提到的方法
|
||||
# class Solution:
|
||||
# def reverseLeftWords(self, s: str, n: int) -> str:
|
||||
# s = list(s)
|
||||
# s[0:n] = list(reversed(s[0:n]))
|
||||
# s[n:] = list(reversed(s[n:]))
|
||||
# s.reverse()
|
||||
class Solution:
|
||||
def reverseLeftWords(self, s: str, n: int) -> str:
|
||||
s = list(s)
|
||||
s[0:n] = list(reversed(s[0:n]))
|
||||
s[n:] = list(reversed(s[n:]))
|
||||
s.reverse()
|
||||
|
||||
return "".join(s)
|
||||
|
||||
# return "".join(s)
|
||||
```
|
||||
|
||||
```python
|
||||
# 方法三：如果连reversed也不让使用，那么自己手写一个
|
||||
class Solution:
|
||||
def reverseLeftWords(self, s: str, n: int) -> str:
|
||||
def reverse_sub(lst, left, right):
|
||||
while left < right:
|
||||
lst[left], lst[right] = lst[right], lst[left]
|
||||
left += 1
|
||||
right -= 1
|
||||
|
||||
res = list(s)
|
||||
end = len(res) - 1
|
||||
reverse_sub(res, 0, n - 1)
|
||||
reverse_sub(res, n, end)
|
||||
reverse_sub(res, 0, end)
|
||||
return ''.join(res)
|
||||
|
||||
# 同方法二
|
||||
# 时间复杂度：O(n)
|
||||
# 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改
|
||||
|
||||
```
|
||||
|
||||
```python 3
|
||||
#方法三：考虑不能用切片的情况下，利用模+下标实现
|
||||
#方法四：考虑不能用切片的情况下，利用模+下标实现
|
||||
class Solution:
|
||||
def reverseLeftWords(self, s: str, n: int) -> str:
|
||||
new_s = ''
|
||||
|
||||
Reference in New Issue
Block a user