2024.5.7 —— LeetCode 高频题复盘

136. 只出现一次的数字

题目链接

Python

方法一

class Solution:def singleNumber(self, nums: List[int]) -> int:ans=nums[0]for i in range(1,len(nums)):ans=nums[i]^ansreturn ans

核心思想：一个数被另一个数异或两次，该数本身不变。

方法二

class Solution:def singleNumber(self, nums: List[int]) -> int:return sum(set(nums))*2-sum(nums)

LCR 155. 将二叉搜索树转化为排序的双向链表

题目链接

Python

"""
# Definition for a Node.
class Node:def __init__(self, val, left=None, right=None):self.val = valself.left = leftself.right = right
"""
class Solution:def treeToDoublyList(self, root: 'Node') -> 'Node':# 1.二叉搜索树——>排序的双向链表(中序遍历)def dfs(cur):if not cur:returndfs(cur.left)# 2. 双向链表（相邻节点）if self.pre:self.pre.right=curcur.left=self.preelse:self.head=curself.pre=curdfs(cur.right)if not root:returnself.pre=Nonedfs(root)# 3. 循环链表（首尾节点）self.head.left=self.preself.pre.right=self.headreturn self.head

11. 盛最多水的容器

题目链接

Python

class Solution:def maxArea(self, height: List[int]) -> int:res = 0i = 0j = len(height) - 1while i < j:area = (j - i) * min(height[i], height[j])res = max(res, area)if height[i] < height[j]:i += 1else:j -= 1return res

498. 对角线遍历

题目链接

Python

class Solution:def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:m,n,ans=len(mat),len(mat[0]),[]for k in range(m+n-1):# 偶对角线（左下到右上）if k%2==0:ans+=[mat[x][k-x] for x in range(min(k,m-1),max(0,k-n+1)-1,-1)]# 奇对角线（右上到左下）else:ans+=[mat[x][k-x] for x in range(max(0,k-n+1),min(k,m-1)+1)]return ans

对角线上横纵坐标和为定值 k 。

402. 移掉 K 位数字

题目链接

Python

class Solution:def removeKdigits(self, num: str, k: int) -> str:stack=[]remain=len(num)-k # 不可少，例如输入为"9",1或者"112",1for n in num:while k and stack and stack[-1]>n:stack.pop()k-=1stack.append(n)return "".join(stack[:remain]).lstrip("0") or "0"

归并排序

Python

class Solution:# 合并两个有序数组def merge(self,left,right):merged=[]i=j=0while i<len(left) and j<len(right):if left[i]<=right[j]:merged.append(left[i])i+=1else:merged.append(right[j])j+=1while i<len(left):merged.append(left[i])i+=1while j<len(right):merged.append(right[j])j+=1return merged# 划分左右数组def sortArray(self, nums: List[int]) -> List[int]:if len(nums)<=1:return numsmid=len(nums)//2left=self.sortArray(nums[:mid])right=self.sortArray(nums[mid:])return self.merge(left,right)

958. 二叉树的完全性检验

题目链接

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def isCompleteTree(self, root: Optional[TreeNode]) -> bool:# 设根节点编号为1，树中所有节点的编号按照广度优先搜索顺序正好是升序，# 我们检测编号序列是否为无间隔的 1, 2, 3, …，事实上，我们只需要检查最后一个编号是否正确，因为最后一个编号的值最大。nodes=[[root,1]]i=0while i<len(nodes):node,v=nodes[i]i+=1if node:nodes.append([node.left,2*v])nodes.append([node.right,2*v+1])return nodes[-1][1]==len(nodes)

123. 买卖股票的最佳时机 III

题目链接

Python

class Solution:def maxProfit(self, prices: List[int]) -> int:# dp[i][0]不操作# dp[i][1]第一次持有股票的最大金额 # dp[i][2]第一次不持有股票的最大金额# dp[i][3]第二次持有股票的最大金额 # dp[i][4]第二次不持有股票的最大金额dp=[[0]*5 for _ in range(len(prices))]dp[0][1]=-prices[0]dp[0][3]=-prices[0]for i in range(1,len(prices)):dp[i][0]=dp[i-1][0]dp[i][1]=max(dp[i-1][1],dp[i-1][0]-prices[i])dp[i][2]=max(dp[i-1][2],dp[i-1][1]+prices[i])dp[i][3]=max(dp[i-1][3],dp[i-1][2]-prices[i])dp[i][4]=max(dp[i-1][4],dp[i-1][3]+prices[i])return dp[-1][-1]

79. 单词搜索

题目链接

Python

class Solution:def exist(self, board: List[List[str]], word: str) -> bool:def dfs(i,j,k):# 不满足条件的进行剪枝if not 0<=i<m or not 0<=j<n or visited[i][j]==True or board[i][j]!=word[k]:return False# 匹配到单词最后一个字符if k==len(word)-1:return Truevisited[i][j]=Truefor d in directions:ni,nj=i+d[0],j+d[1]if dfs(ni,nj,k+1):return Truevisited[i][j]=Falsem,n=len(board),len(board[0])directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]visited=[[False for _ in range(n)] for _ in range(m)]for i in range(m):for j in range(n):if dfs(i,j,0):return Truereturn False

排序奇升偶降链表

题目链接

在做该题之前建议先做 206. 反转链表、21. 合并两个有序链表、328. 奇偶链表这三题。

Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:    def sortOddEvenList(self,head):     if not head or not head.next:      return head        oddList,evenList = self.partition(head)        evenList = self.reverse(evenList)        return self.merge(oddList,evenList)    def partition(self, head):        evenHead = head.next        odd, even = head, evenHead        while even and even.next:            odd.next = even.next            odd = odd.next            even.next = odd.next            even = even.next        odd.next = None        return [head,evenHead] def reverse(self, head):pre=Nonecur=headwhile cur:temp=cur.nextcur.next=prepre=curcur=tempreturn predef merge(self, list1, list2):if not list1:return list2if not list2:return list1if list1.val<list2.val:list1.next=self.merge(list1.next,list2)return list1else:list2.next=self.merge(list2.next,list1)return list2