LeetCode 每日一题 2025/2/3-2025/2/9

记录了初步解题思路 以及本地实现代码；并不一定为最优 也希望大家能一起探讨 一起进步

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2/3 680. 验证回文串 II

头尾比较 可以有一次误差 遇到不一致时
分两种情况 去掉left或者去掉right 其中一种满足就可

def validPalindrome(s):""":type s: str:rtype: bool"""def sub(s):if len(s)==0 or len(s)==1:return Truei,j=0,len(s)-1while i<j:if s[i]!=s[j]:return Falseelse:i+=1j-=1return Truei,j=0,len(s)-1while i<j:if s[i]!=s[j]:return sub(s[i+1:j+1]) or sub(s[i:j])           else:i+=1j-=1return Truedef validPalindrome2(s):""":type s: str:rtype: bool"""if s==s[::-1]:return Truei,j=0,len(s)-1while i<j:if s[i]==s[j]:i+=1j-=1else:tmp = s[:i]+s[i+1:]if tmp==tmp[::-1]:return Truetmp = s[:j]+s[j+1:]if tmp==tmp[::-1]:return Truereturn False

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2/4 922. 按奇偶排序数组 II

1. 分奇数偶数两个list 在偶数list中的奇数位 插入奇数
2. 用ij记录偶数奇数位置 遍历list 如果是偶数放入i位置 奇数放入j位置
3. 遍历一次数组
   使用ji,ou两个数组记录不符合未知的数
   遇到不符合位置的数时 从ji,ou数组中找到时候存在可交换的位置 有则交换 无则将该位置放入数组

def sortArrayByParityII(nums):""":type nums: List[int]:rtype: List[int]"""odd = [x for x in nums if x%2==1]ans = [x for x in nums if x%2==0]p = 1for i in odd:ans.insert(p,i)p+=2return ansdef sortArrayByParityII2(nums):""":type nums: List[int]:rtype: List[int]"""i,j=0,1ans = [0]*len(nums)for x in nums:if x%2==0:ans[i]=xi+=2else:ans[j]=xj+=2def sortArrayByParityII3( nums):""":type nums: List[int]:rtype: List[int]"""ji=[]ou=[]for i in range(len(nums)):if i%2!=nums[i]%2:if i%2==0:if len(ji)>0:loc = ji.pop()nums[loc],nums[i]=nums[i],nums[loc]else:ou.append(i)else:if len(ou)>0:loc = ou.pop()nums[loc],nums[i]=nums[i],nums[loc]else:ji.append(i)return nums

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2/5 90. 子集 II

set 记录重复

def subsetsWithDup(nums):""":type nums: List[int]:rtype: List[List[int]]"""ret = []ret.append([])ck = set()for num in nums:l = []for tmp in ret:x = tmp[:]x.append(num)t = tuple(sorted(x))if t not in ck:l.append(x)ck.add(t)     ret.extend(l)return ret

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2/6 47. 全排列 II

统计每个数字出现次数 回溯

def permuteUnique(nums):""":type nums: List[int]:rtype: List[List[int]]"""m={}for i in nums:m[i] = m.get(i,0)+1n = len(nums)ret = []def back(m,l):if len(l)==n:ret.append(l)returnfor c in m:if m[c]>0:m[c]-=1l.append(c)back(m,l[:])l.pop()m[c]+=1back(m,[])return ret

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2/7 59. 螺旋矩阵 II

顺指针可以 → ↓ ← ↑ 分别对应四种坐标变化(0,1),(1,0),(0,-1),(-1,0)
从→开始 如果遇到边界或者重复则改为下一种走法

def generateMatrix(n):""":type n: int:rtype: List[List[int]]"""matrix = [[0]*n for _ in range(n)]steplist =[(0,1),(1,0),(0,-1),(-1,0)]step = 0num=2i,j=0,0matrix[0][0]=1while num<=n*n:x,y = steplist[step]ti,tj = i+x,j+yif ti<0 or tj<0 or ti>=n or tj>=n or matrix[ti][tj]>0:step = (step+1)%4continuei,j = ti,tjmatrix[i][j] = numnum+=1return matrix

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2/8 63. 不同路径 II

抵达当前位置的路径数量为 左侧位置和上方位置数量之和

def uniquePathsWithObstacles(obstacleGrid):""":type obstacleGrid: List[List[int]]:rtype: int"""m,n=len(obstacleGrid),len(obstacleGrid[0])ans = [[0]*n for _ in range(m)]for i in range(m):if obstacleGrid[i][0]==0:ans[i][0]=1else:breakfor j in range(n):if obstacleGrid[0][j]==0:ans[0][j]=1else:breakfor i in range(1,m):for j in range(1,n):if obstacleGrid[i][j]==0:ans[i][j]=ans[i-1][j]+ans[i][j-1]return ans[-1][-1]

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2/9 80. 删除有序数组中的重复项 II

1.使用map记录重复
2.查看与前两位的比较 是否重复两次以上

def removeDuplicates(nums):""":type nums: List[int]:rtype: int"""m={}loc =0for i in range(len(nums)):if m.get(nums[i],0)>=2:continuem[nums[i]] = m.get(nums[i],0)+1nums[loc] = nums[i]loc+=1return locdef removeDuplicates2(nums):""":type nums: List[int]:rtype: int"""if len(nums)<3:return len(nums)loc =2for i in range(2,len(nums)):if nums[loc-2]==nums[i]:continuenums[loc] = nums[i]loc+=1return loc

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