简单题
两数之和
方法一,暴力破解,时间复杂度O(n^2),空间复杂度O(1)
class Solution:def twoSum(self, nums: List[int], target: int) -> List[int]:n=len(nums)for i in range(n):for j in range(i+1,n):if nums[i]+nums[j]==target:return[i,j]
方法二,哈希表,时间复杂度O(n),空间复杂度O(n)
class Solution:def twoSum(self, nums: List[int], target: int) -> List[int]:h={}# 向右枚举j,将值和下标存入哈希表,然后再枚举左边的数i,找到对应下标for j,x in enumerate(nums):if target-x in h:return [h[target-x],j] h[x]=j
移动零
方法一:把非零数都移到前面,其他数都赋值为0
class Solution:def moveZeroes(self, nums: List[int]) -> None:"""Do not return anything, modify nums in-place instead."""i=0for j in range(len(nums)):if nums[j]!=0:nums[i]=nums[j]i+=1for x in range(i,len(nums)):nums[x]=0return nums
方法二: 双指针
class Solution:def moveZeroes(self, nums: List[int]) -> None:"""Do not return anything, modify nums in-place instead."""l=0for r in range(len(nums)):if nums[r]!=0:nums[l],nums[r]=nums[r],nums[l]l+=1return nums
反转字符串
双指针
class Solution:def reverseString(self, s: List[str]) -> None:"""Do not return anything, modify s in-place instead."""l,r=0,len(s)-1while l<r:s[l],s[r]=s[r],s[l]l+=1r-=1return s
同构字符串
map方法
zip方法
index方法
class Solution:def isIsomorphic(self, s: str, t: str) -> bool:for i in range(len(s)):if s.index(s[i]) != t.index(t[i]):return Falsereturn True
反转链表
时间复杂度O(n),空间复杂度O(1)
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:pre=Nonecur=headwhile cur:temp=cur.nextcur.next=prepre=curcur=tempreturn pre
字符串相加
双指针,让两个指针指向尾部,模拟人工加法,处理好进位,然后再拼接起来
class Solution:def addStrings(self, num1: str, num2: str) -> str:res=""i,j,carry=len(num1)-1,len(num2)-1,0while i>=0 or j >=0:# 超出索引范围直接赋值为0,相当于长度短的数字前面填0,方便后续计算n1=int(num1[i]) if i>=0 else 0n2=int(num2[j]) if j>=0 else 0temp=n1+n2+carrycarry = temp//10res=str(temp%10)+resi,j=i-1,j-1# 循环结束时,如果carry为1,则加到头部,没有则直接返回结果return "1"+res if carry else res
找出字符串中第一个匹配项的下标
方法一:直接用Python的内置函数index或者find,可以用来寻找字符串中子串,如果能找到,则返回第一个下标,否则抛出异常。
时间复杂度O(n)
class Solution:def strStr(self, haystack: str, needle: str) -> int:try:return haystack.index(needle)except:return -1
方法二:切片
class Solution:def strStr(self, haystack: 'str', needle: 'str') -> 'int':for i in range(0, len(haystack) - len(needle) + 1):if haystack[i:i+len(needle)] == needle:return ireturn -1
最长和谐子序列
方法一:哈希表存储,用HashMap统计个数后,遍历的时候直接判断有没有相邻的数的个数统计答案即可
class Solution:def findLHS(self, nums: List[int]) -> int:cnt = defaultdict(int)ans=0for num in nums:cnt[num]+=1if num+1 in cnt:ans=max(ans,cnt[num]+cnt[num+1])if num-1 in cnt:ans=max(ans,cnt[num]+cnt[num-1])return ans
方法二:排序+双指针时间复杂度O(nlog(n)),空间复杂度O(1)
class Solution:def findLHS(self, nums: List[int]) -> int:nums.sort()left=0res=0for right in range(len(nums)):if nums[right]-nums[left]>1:left+=1if nums[right]-nums[left]==1:res=max(res,right-left+1)return res
存在重复元素
方法一:哈希集合,将数组转化为集合(会去掉重复元素),然后判断前后长度是否一致
class Solution:def containsDuplicate(self, nums: List[int]) -> bool:s=set(nums)if len(s)!=len(nums):return Trueelse:return False
方法二:哈希表
class Solution:def containsDuplicate(self, nums: List[int]) -> bool:numset={}for i in nums:if i not in numset:numset[i]=1else:return Truereturn False
字符串中的第一个唯一字符
方法一:Counter法,统计字符串中每一个字符出现次数,然后遍历到第一个字符直接返回下标
class Solution:def firstUniqChar(self, s: str) -> int:l=Counter(s)for i in s:if l[i]==1:return s.index(i)return -1
方法二:哈希表,用哈希表存储每个字符,如果出现次数大于1则值为False,等于1则为True
class Solution:def firstUniqChar(self, s: str) -> int:h={}for i in s:h[i]=not i in hfor j,x in enumerate(s):if h[c]:return ifreturn -1
罗马数字转整数
删除排序链表中的重复元素
快慢双指针,让快指针在前面探路,发现不重复元素,就放到慢指针后面,慢指针就往前一步,一直到遍历完整张表。此时从head到slow都是不重复元素,再断开slow后面的指针,即可去重。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:slow,fast=head,headif head is None:return Nonewhile fast:if fast.val!=slow.val:slow.next=fastslow=slow.nextfast=fast.nextslow.next=Nonereturn head
删除排序数组中的重复元素
快慢指针,同上一题
class Solution:def removeDuplicates(self, nums: List[int]) -> int:if len(nums)==0:return 0slow,fast=0,0while fast<len(nums):if nums[fast]!=nums[slow]:slow+=1nums[slow]=nums[fast]fast+=1return slow+1
最长公共前缀
方法一:Python内置函数zip,可以把多个序列中的元素打包成一个元组,zip(*可迭代对象)
class Solution:def longestCommonPrefix(self, strs: List[str]) -> str:l=0for i in zip(*strs):if len(set(i))>1:breakl+=1return strs[0][:l]
有效的括号
括号匹配,用栈来做最佳
方法一:栈+if-else循环
时间复杂度O(n),空间复杂度是O(n)
class Solution:def isValid(self, s: str) -> bool:if len(s)%2!=0:return Falsestk=[]for c in s:if c=="(":stk.append(")")elif c=="[":stk.append("]")elif c=="{":stk.append("}")elif not stk or stk.pop()!=c:return Falsereturn not stk
方法二:栈+哈希表
class Solution:def isValid(self, s: str) -> bool:if len(s)%2!=0:return Falsestk=[]mp={'(':')','[':']','{':'}'}for c in s:if c in mp:stk.append(mp[c])elif not stk or stk.pop()!=c:return Falsereturn not stk
反转字符串中的元音字母
对撞双指针,将元音字母顺序交换
class Solution:def reverseVowels(self, s: str) -> str:# a e i o u #元音元素dic = {'a','e','i','o','u','A','E','I','O','U'} #大小写元音元素集合作为判断依据pol = 0 #左指针por = len(s)-1 #右指针s_ = list(s) #str类型数据无法直接查询in和not in,转换为listwhile pol < por: #左右指针交错循环停止if s_[pol] in dic and s_[por] in dic: #左右指针所指元素均在集合中s_[pol], s_[por] = s_[por], s_[pol] #交换左右指针所指元素por -= 1 #右指针左移pol += 1 #左指针右移if s_[por] not in dic: #右指针所指元素不在集合中por -= 1 #右指针左移if s_[pol] not in dic: #左指针所指元素不在集合中pol += 1 #左指针右移return ''.join(s_) #返回str格式数据
二进制转十进制
def binary_to_decimal(binary_array):# 将数组转换成字符串binary_str = ''.join(str(bit) for bit in binary_array)# 判断是否为负数is_negative = binary_str[0] == '1'if is_negative:# 对于负数,转换为原码后计算inverted = ['0' if b == '1' else '1' for b in binary_str]inverted_str = ''.join(inverted)# 加1操作decimal_value = -int(inverted_str, 2) + 1else:# 正数直接转换decimal_value = int(binary_str, 2)return decimal_value# 示例
binary_array = [1, 1, 1, 1] # 表示-1(假设字长为4)
print(binary_to_decimal(binary_array)) # 输出: -1
中等题
三数之和
先给数组排序,固定好一个值,再创建两个对撞指针,遍历整张表,然后删除重复的元素。
class Solution:def threeSum(self, nums: List[int]) -> List[List[int]]:nums.sort()res = []n=len(nums)for i in range(n-2):x=nums[i]if i>0 and x==nums[i-1]:continuel=i+1r=n-1while l<r:s=x+nums[l]+nums[r]if s<0:l+=1elif s>0:r-=1else:res.append([x,nums[l],nums[r]])l+=1# 删除重复元素while l<r and nums[l]==nums[l-1]:l+=1r-=1while l<r and nums[r]==nums[r+1]:r-=1return res
反转字符串中的单词
class Solution:def reverseWords(self, s: str) -> str:# 删除首尾空格s=s.strip()# 分割字符串strs=s.split()# 翻转单词列表strs.reverse()return " ".join(strs)
两数相加
# 注意:python 代码由 chatGPT🤖 根据我的 java 代码翻译。
# 本代码的正确性已通过力扣验证,如有疑问,可以对照我的 java 代码查看。class Solution:def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:# 在两条链表上的指针p1, p2 = l1, l2# 虚拟头结点(构建新链表时的常用技巧)dummy = ListNode(-1)# 指针 p 负责构建新链表p = dummy# 记录进位carry = 0# 开始执行加法,两条链表走完且没有进位时才能结束循环while p1 is not None or p2 is not None or carry > 0:# 先加上上次的进位val = carryif p1 is not None:val += p1.valp1 = p1.nextif p2 is not None:val += p2.valp2 = p2.next# 处理进位情况carry = val // 10val = val % 10# 构建新节点p.next = ListNode(val)p = p.next# 返回结果链表的头结点(去除虚拟头结点)return dummy.next
无重复字符的最长子串
滑动窗口
class Solution:def lengthOfLongestSubstring(self, s: str) -> int:ans=0l=0h=Counter()for r,x in enumerate(s):h[x]+=1while h[x]>1:h[s[l]]-=1l+=1ans=max(ans,r-l+1)return ans
每日温度
class Solution:def dailyTemperatures(self, temperatures):n = len(temperatures)res = [0]*n# 这里放元素索引,而不是元素s = []# 单调栈模板for i in range(n-1, -1, -1):while s and temperatures[s[-1]] <= temperatures[i]:s.pop()# 得到索引间距res[i] = 0 if not s else s[-1] - i# 将索引入栈,而不是元素s.append(i)return res
删除排序链表中重复元素II
class Solution:def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:cur = dummy = ListNode(next=head)while cur.next and cur.next.next:val = cur.next.valif cur.next.next.val == val:while cur.next and cur.next.val == val:cur.next = cur.next.nextelse:cur = cur.nextreturn dummy.next
跳跃游戏
class Solution:def canJump(self, nums: List[int]) -> bool:max_i=0for i,jump in enumerate(nums):if max_i>=i and i+jump>max_i:max_i=i+jumpreturn max_i>=i
和为K的子数组
class Solution:def subarraySum(self, nums: List[int], k: int) -> int:s = [0] * (len(nums) + 1)for i, x in enumerate(nums):s[i + 1] = s[i] + xans = 0cnt = defaultdict(int)for sj in s:ans += cnt[sj - k]cnt[sj] += 1return ans
买卖股票的最佳时机
class Solution:def maxProfit(self, prices: List[int]) -> int:ans = 0min_pirce = prices[0]for p in prices:ans = max(ans, p - min_pirce)min_pirce = min(min_pirce, p)return ans
整数转罗马数字
R = (("", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"), # 个位("", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"), # 十位("", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"), # 百位("", "M", "MM", "MMM"), # 千位
)class Solution:def intToRoman(self, num: int) -> str:return R[3][num // 1000] + R[2][num // 100 % 10] + R[1][num // 10 % 10] + R[0][num % 10]