diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md index a26aa308..57729c38 100644 --- a/problems/0020.有效的括号.md +++ b/problems/0020.有效的括号.md @@ -162,18 +162,44 @@ class Solution { Python： ```python3 +# 方法一，仅使用栈，更省空间 class Solution: def isValid(self, s: str) -> bool: - stack = [] # 保存还未匹配的左括号 - mapping = {")": "(", "]": "[", "}": "{"} - for i in s: - if i in "([{": # 当前是左括号，则入栈 - stack.append(i) - elif stack and stack[-1] == mapping[i]: # 当前是配对的右括号则出栈 - stack.pop() - else: # 不是匹配的右括号或者没有左括号与之匹配，则返回false + stack = [] + + for item in s: + if item == '(': + stack.append(')') + elif item == '[': + stack.append(']') + elif item == '{': + stack.append('}') + elif not stack or stack[-1] != item: return False - return stack == [] # 最后必须正好把左括号匹配完 + else: + stack.pop() + + return True if not stack else False +``` + +```python3 +# 方法二，使用字典 +class Solution: + def isValid(self, s: str) -> bool: + stack = [] + mapping = { + '(': ')', + '[': ']', + '{': '}' + } + for item in s: + if item in mapping.keys(): + stack.append(mapping[item]) + elif not stack or stack[-1] != item: + return False + else: + stack.pop() + return True if not stack else False ``` Go： diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md index 91e566dd..672b9a8f 100644 --- a/problems/0024.两两交换链表中的节点.md +++ b/problems/0024.两两交换链表中的节点.md @@ -160,21 +160,29 @@ class Solution { Python： ```python +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, val=0, next=None): +# self.val = val +# self.next = next + class Solution: def swapPairs(self, head: ListNode) -> ListNode: - dummy = ListNode(0) #设置一个虚拟头结点 - dummy.next = head - cur = dummy - while cur.next and cur.next.next: - tmp = cur.next #记录临时节点 - tmp1 = cur.next.next.next #记录临时节点 + res = ListNode(next=head) + pre = res + + # 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 + while pre.next and pre.next.next: + cur = pre.next + post = pre.next.next - cur.next = cur.next.next #步骤一 - cur.next.next = tmp #步骤二 - cur.next.next.next = tmp1 #步骤三 - - cur = cur.next.next #cur移动两位，准备下一轮交换 - return dummy.next + # pre，cur，post对应最左，中间的，最右边的节点 + cur.next = post.next + post.next = cur + pre.next = post + + pre = pre.next.next + return res.next ``` Go： diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index 2b294337..bcde7d5b 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -223,17 +223,19 @@ var evalRPN = function(tokens) { python3 ```python -def evalRPN(tokens) -> int: - stack = list() - for i in range(len(tokens)): - if tokens[i] not in ["+", "-", "*", "/"]: - stack.append(tokens[i]) - else: - tmp1 = stack.pop() - tmp2 = stack.pop() - res = eval(tmp2+tokens[i]+tmp1) - stack.append(str(int(res))) - return stack[-1] +class Solution: + def evalRPN(self, tokens: List[str]) -> int: + stack = [] + for item in tokens: + if item not in {"+", "-", "*", "/"}: + stack.append(item) + else: + first_num, second_num = stack.pop(), stack.pop() + stack.append( + int(eval(f'{second_num} {item} {first_num}')) # 第一个出来的在运算符后面 + ) + return int(stack.pop()) # 如果一开始只有一个数，那么会是字符串形式的 + ``` diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index f3724fc2..d67a7d2a 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -245,13 +245,15 @@ Python： # def __init__(self, val=0, next=None): # self.val = val # self.next = next + class Solution: def removeElements(self, head: ListNode, val: int) -> ListNode: - dummy_head = ListNode(next=head) #添加一个虚拟节点 + dummy_head = ListNode(next=head) cur = dummy_head - while(cur.next!=None): - if(cur.next.val == val): - cur.next = cur.next.next #删除cur.next节点 + + while cur.next: + if cur.next.val == val: + cur.next = cur.next.next # 删除下一个节点 else: cur = cur.next return dummy_head.next diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index b327c17b..ccf93f1f 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -294,53 +294,66 @@ Python： ```python from collections import deque + class MyStack: + def __init__(self): """ - Initialize your data structure here. + Python普通的Queue或SimpleQueue没有类似于peek的功能 + 也无法用索引访问，在实现top的时候较为困难。 + + 用list可以，但是在使用pop(0)的时候时间复杂度为O(n) + 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 + + in - 存所有数据 + out - 仅在pop的时候会用到 """ - #使用两个队列来实现 - self.que1 = deque() - self.que2 = deque() + self.queue_in = deque() + self.queue_out = deque() def push(self, x: int) -> None: """ - Push element x onto stack. + 直接append即可 """ - self.que1.append(x) + self.queue_in.append(x) + def pop(self) -> int: """ - Removes the element on top of the stack and returns that element. + 1. 首先确认不空 + 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 + 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 + 4. 交换in和out，此时out里只有一个元素 + 5. 把out中的pop出来，即是原队列的最后一个 + + 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 + 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 """ - size = len(self.que1) - size -= 1#这里先减一是为了保证最后面的元素 - while size > 0: - size -= 1 - self.que2.append(self.que1.popleft()) + if self.empty(): + return None - - result = self.que1.popleft() - 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 - #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 - return result + for i in range(len(self.queue_in) - 1): + self.queue_out.append(self.queue_in.popleft()) + + 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 + return self.queue_out.popleft() def top(self) -> int: """ - Get the top element. + 1. 首先确认不空 + 2. 我们仅有in会存放数据，所以返回第一个即可 """ - return self.que1[-1] + if self.empty(): + return None + + return self.queue_in[-1] + def empty(self) -> bool: """ - Returns whether the stack is empty. + 因为只有in存了数据，只要判断in是不是有数即可 """ - #print(self.que1) - if len(self.que1) == 0: - return True - else: - return False - + return len(self.queue_in) == 0 ``` diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md index 0df82d35..27a3de4e 100644 --- a/problems/0232.用栈实现队列.md +++ b/problems/0232.用栈实现队列.md @@ -281,48 +281,60 @@ class MyQueue { Python： ```python -# 使用两个栈实现先进先出的队列 class MyQueue: + def __init__(self): """ - Initialize your data structure here. + in主要负责push，out主要负责pop """ - self.stack1 = list() - self.stack2 = list() + self.stack_in = [] + self.stack_out = [] + def push(self, x: int) -> None: """ - Push element x to the back of queue. + 有新元素进来，就往in里面push """ - # self.stack1用于接受元素 - self.stack1.append(x) + self.stack_in.append(x) + def pop(self) -> int: """ - Removes the element from in front of queue and returns that element. + 1. 检查如果out里面元素，则直接pop + 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 + 3. 直接把in剩下的元素pop出来，就是queue头部的 """ - # 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 - if self.stack2 == []: - while self.stack1: - tmp = self.stack1.pop() - self.stack2.append(tmp) - return self.stack2.pop() + if self.empty: + return None + + if self.stack_out: + return self.stack_out.pop() + else: + for i in range(1, len(self.stack_in)): + self.stack_out.append(self.stack_in.pop()) + return self.stack_in.pop() + def peek(self) -> int: """ - Get the front element. + 1. 查out有没有元素，有就把最上面的返回 + 2. 如果out没有元素，就把in最下面的返回 """ - if self.stack2 == []: - while self.stack1: - tmp = self.stack1.pop() - self.stack2.append(tmp) - return self.stack2[-1] + if self.empty: + return None + + if self.stack_out: + return self.stack_out[-1] + else: + return self.stack_in[0] + def empty(self) -> bool: """ - Returns whether the queue is empty. + 只要in或者out有元素，说明队列不为空 """ - return self.stack1 == [] and self.stack2 == [] + return not (self.stack_in or self.stack_out) + ``` diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 4f96f839..763097be 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -162,21 +162,14 @@ class Solution: Do not return anything, modify s in-place instead. """ left, right = 0, len(s) - 1 - while(left < right): + + # 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 + # 推荐该写法，更加通俗易懂 + while left < right: s[left], s[right] = s[right], s[left] left += 1 right -= 1 - -# 下面的写法更加简洁，但是都是同样的算法 -# class Solution: -# def reverseString(self, s: List[str]) -> None: -# """ -# Do not return anything, modify s in-place instead. -# """ - # 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): -# s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i] -# return s + ``` Go： diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index ab1ef16a..02713c65 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -155,34 +155,27 @@ class Solution { Python： ```python - -class Solution(object): - def reverseStr(self, s, k): +class Solution: + def reverseStr(self, s: str, k: int) -> str: """ - :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) ``` diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md index c4ae85c9..b60a8d1d 100644 --- a/problems/1047.删除字符串中的所有相邻重复项.md +++ b/problems/1047.删除字符串中的所有相邻重复项.md @@ -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： diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 4ae5f9f2..3f373d0d 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -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) ``` diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md index 1073dafa..15607a50 100644 --- a/problems/剑指Offer58-II.左旋转字符串.md +++ b/problems/剑指Offer58-II.左旋转字符串.md @@ -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 = ''