随笔档案「2020年4月」 - Sempron2800+

leetcode1408

摘要：1 class Solution: 2 def stringMatching(self, words: 'List[str]') -> 'List[str]': 3 n = len(words) 4 words = sorted(words,key=lambda x:len(x)) 5 res = 阅读全文

posted @ 2020-04-12 12:41 Sempron2800+ 阅读(185) 评论(0) 推荐(0)

leetcode396

摘要：C++的实现： 1 class Solution { 2 public: 3 int maxRotateFunction(vector<int>& A) { 4 long N = A.size(); 5 long S = 0; 6 long t = 0; 7 for (int i = 0; i < 阅读全文

posted @ 2020-04-09 10:43 Sempron2800+ 阅读(193) 评论(0) 推荐(0)

leetcode393

摘要：1 class Solution: 2 def validUtf8(self, data): 3 # 标记这个字节是某个编码的第几个字节 4 n_bytes = 0 5 6 # 遍历数组 7 for num in data: 8 9 # 获取二进制编码，保留最低8位 10 bin_rep = for 阅读全文

posted @ 2020-04-09 10:42 Sempron2800+ 阅读(208) 评论(0) 推荐(0)

leetcode390

摘要：1 class Solution { 2 public: 3 int lastRemaining(int n) { 4 if (n == 1) return 1; 5 return 2 * (n / 2 + 1 - lastRemaining(n / 2)); 6 7 // vector<int> 阅读全文

posted @ 2020-04-09 10:40 Sempron2800+ 阅读(223) 评论(0) 推荐(0)

leetcode388

摘要：1 class Solution(object): 2 def lengthLongestPath(self, input): 3 """ 4 :type input: str 5 :rtype: int 6 """ 7 input = input.split('\n') 8 res = 0 9 s 阅读全文

posted @ 2020-04-09 10:34 Sempron2800+ 阅读(166) 评论(0) 推荐(0)

leetcode373

摘要：1 class Solution: 2 def kSmallestPairs(self, nums1, nums2, k): 3 queue = [] 4 def push(i, j): 5 if i < len(nums1) and j < len(nums2): 6 heapq.heappush 阅读全文

posted @ 2020-04-09 10:30 Sempron2800+ 阅读(237) 评论(0) 推荐(0)

leetcode368

摘要：1 class Solution(object): 2 def largestDivisibleSubset(self, nums): 3 """ 4 :type nums: List[int] 5 :rtype: List[int] 6 """ 7 # The container that hol 阅读全文

posted @ 2020-04-09 10:27 Sempron2800+ 阅读(139) 评论(0) 推荐(0)

leetcode372

摘要：1 class Solution: 2 def superPow(self, a: int, b: List[int]) -> int: 3 return pow(a,int(''.join(map(str,b))),1337) 算法思路：直接调用python的pow函数。学有余力的同学，可以尝试阅读全文

posted @ 2020-04-09 10:18 Sempron2800+ 阅读(135) 评论(0) 推荐(0)

leetcode386

摘要：1 class Solution: 2 def lexicalOrder(self, n: int) -> List[int]: 3 lis = [str(i) for i in range(1,n+1)] 4 lis.sort() 5 return lis 算法思路：按字符串顺序排序。当然有能力阅读全文

posted @ 2020-04-09 10:12 Sempron2800+ 阅读(106) 评论(0) 推荐(0)

leetcode385

摘要：1 class Solution: 2 def deserialize(self, s: str) -> NestedInteger: 3 4 if s[0] != '[': 5 return NestedInteger(int(s)) 6 7 stack = [] 8 # num为数字，sign为阅读全文

posted @ 2020-04-09 09:44 Sempron2800+ 阅读(174) 评论(0) 推荐(0)

leetcode397

摘要：1 class Solution: 2 def integerReplacement(self, n: int) -> int: 3 count = 0 4 while n != 1: 5 if (n & 1) == 0: # 偶数直接右移 6 n >>= 1 7 else: 8 n += -1 i 阅读全文

posted @ 2020-04-09 09:41 Sempron2800+ 阅读(135) 评论(0) 推荐(0)

leetcode365

摘要：方法一：深度优先搜索 1 class Solution: 2 def canMeasureWater(self, x: int, y: int, z: int) -> bool: 3 stack = [(0, 0)] 4 self.seen = set() 5 while stack: 6 rema 阅读全文

posted @ 2020-04-09 09:30 Sempron2800+ 阅读(169) 评论(0) 推荐(0)

leetcode355

摘要：1 import heapq 2 import collections 3 class Twitter: 4 def __init__(self): 5 self.followers = collections.defaultdict(set)#key是被关注者，value是关注这个用户的人的集合阅读全文

posted @ 2020-04-09 08:54 Sempron2800+ 阅读(232) 评论(0) 推荐(0)

leetcode332

摘要：没能做出来，参考别人的： 1 from heapq import heapify, heappush, heappop 2 3 class Solution: 4 def findItinerary(self, tickets: List[List[str]]) -> List[str]: 5 gr 阅读全文

posted @ 2020-04-08 20:18 Sempron2800+ 阅读(193) 评论(0) 推荐(0)

leetcode331

摘要：1 class Solution: 2 def isValidSerialization(self, preorder: str) -> bool: 3 if preorder == '#':#空树是合法的 4 return True 5 nodes = preorder.split(',') 6 阅读全文

posted @ 2020-04-08 17:54 Sempron2800+ 阅读(184) 评论(0) 推荐(0)

leetcode313

摘要：1 import sys 2 class Solution: 3 def nthSuperUglyNumber(self, n: int, primes: 'List[int]') -> int: 4 dp = [1]*n 5 m = len(primes) 6 plist = [0] * m 7 阅读全文

posted @ 2020-04-08 16:15 Sempron2800+ 阅读(161) 评论(0) 推荐(0)

leetcode310

摘要：尝试使用bfs解决，但是TLE。给出代码如下： 1 import sys 2 class Solution: 3 def __init__(self): 4 self.depth = [] 5 self.mindepth = sys.maxsize 6 7 def bfs(self,n,visite 阅读全文

posted @ 2020-04-08 11:53 Sempron2800+ 阅读(160) 评论(0) 推荐(0)

leetcode306

摘要：1 class Solution: 2 def backTrack(self,num,l,idx,n): 3 if idx >= n: 4 if len(l) >= 3: 5 return True 6 else: 7 return False 8 for i in range(idx,n): 9 阅读全文

posted @ 2020-04-07 21:49 Sempron2800+ 阅读(181) 评论(0) 推荐(0)

leetcode307

摘要：1 class segTree: 2 def __init__(self,nums): 3 n = len(nums) 4 for i in range(1,31): 5 low,high = 2 ** (i-1),2 ** i 6 if n > low and n < high: 7 fillze 阅读全文

posted @ 2020-04-07 13:40 Sempron2800+ 阅读(160) 评论(0) 推荐(0)

leetcode304

摘要：1 class NumMatrix: 2 def __init__(self, matrix: 'List[List[int]]'): 3 self.matrix = matrix 4 m,n = 0,0 5 m = len(matrix) 6 if m > 0: 7 n = len(matrix[ 阅读全文

posted @ 2020-04-07 09:41 Sempron2800+ 阅读(158) 评论(0) 推荐(0)

leetcode45

摘要：1 class Solution(object): 2 def jump(self, nums): 3 """ 4 :type nums: List[int] 5 :rtype: int 6 """ 7 ln = len(nums) 8 curr = 0 9 last = 0 10 step = 0 阅读全文

posted @ 2020-04-06 10:10 Sempron2800+ 阅读(166) 评论(0) 推荐(0)

leetcode1405

摘要：第一种方案，使用堆： 1 from heapq import heappush, heappop 2 class Solution: 3 def longestDiverseString(self, a: int, b: int, c: int) -> str: 4 max_heap = [] 5 阅读全文

posted @ 2020-04-06 08:50 Sempron2800+ 阅读(257) 评论(0) 推荐(0)

leecode220

摘要：1 class Solution: 2 def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool: 3 if t == 0 and len(set(nums)) == len(nums): 4 r 阅读全文

posted @ 2020-04-06 08:48 Sempron2800+ 阅读(119) 评论(0) 推荐(0)

leetcode275

摘要：1 class Solution: 2 def hIndex(self, citations): 3 citations_len = len(citations) 4 if citations_len<=0: 5 return 0 6 low = 0 7 high = citations_len-1 阅读全文

posted @ 2020-04-06 08:44 Sempron2800+ 阅读(168) 评论(0) 推荐(0)

leetcode274

摘要：1 class Solution: 2 def hIndex(self, citations: 'List[int]') -> int: 3 n = len(citations) 4 if n == 0: 5 return 0 6 citations = sorted(citations,rever 阅读全文

posted @ 2020-04-06 08:41 Sempron2800+ 阅读(107) 评论(0) 推荐(0)

leetcode284

摘要：1 class PeekingIterator: 2 def __init__(self, iterator): 3 self.iterator = iterator 4 self.head = iterator.next() 5 6 def peek(self): 7 return self.he 阅读全文

posted @ 2020-04-06 08:16 Sempron2800+ 阅读(144) 评论(0) 推荐(0)

leetcode1404

摘要：1 class Solution: 2 def convertInt(self,s): 3 n = len(s) 4 basenum = 0 5 p = 0 6 for i in range(n-1,-1,-1): 7 basenum += int(s[i]) * (2 ** p) 8 p += 1 阅读全文

posted @ 2020-04-05 12:57 Sempron2800+ 阅读(179) 评论(0) 推荐(0)

leetcode1403

摘要：1 class Solution: 2 def minSubsequence(self, nums: 'List[int]') -> 'List[int]': 3 nums = sorted(nums,reverse=True) 4 n = len(nums) 5 if n == 1: 6 retu 阅读全文

posted @ 2020-04-05 12:55 Sempron2800+ 阅读(170) 评论(0) 推荐(0)

leetcode264

摘要：先给出一个我自己写的方案，TLE 1 class Solution: 2 def __init__(self): 3 self.memo = dict() 4 self.memo[1] = True 5 6 def isUgly(self,num): 7 if num <= 0: 8 return 阅读全文

posted @ 2020-04-05 09:53 Sempron2800+ 阅读(220) 评论(0) 推荐(0)

leetcode229

摘要：1 class Solution: 2 def majorityElement(self, nums: List[int]) -> List[int]: 3 dic = dict() 4 n = len(nums) 5 for i in range(n): 6 if nums[i] not in d 阅读全文

posted @ 2020-04-05 09:25 Sempron2800+ 阅读(136) 评论(0) 推荐(0)

leetcode1401

摘要：1 class Solution: 2 def checkOverlap(self, radius: int, x_center: int, y_center: int, x1: int, y1: int, x2: int, y2: int) -> bool: 3 4 # Getting the c 阅读全文

posted @ 2020-04-05 05:05 Sempron2800+ 阅读(154) 评论(0) 推荐(0)

leetcode1400

摘要：1 class Solution: 2 def canConstruct(self, s: str, k: int) -> bool: 3 #奇数个字符出现的数量 <= k 4 #所有的字符的类别数量 >= k 5 dic = dict() 6 n = len(s) 7 for i in range 阅读全文

posted @ 2020-04-05 05:00 Sempron2800+ 阅读(142) 评论(0) 推荐(0)

leetcode1399

摘要：1 class Solution: 2 def calSum(self,x): 3 s = str(x) 4 sums = 0 5 for j in range(len(s)): 6 sums += int(s[j]) 7 return sums 8 9 def countLargestGroup( 阅读全文

posted @ 2020-04-05 04:55 Sempron2800+ 阅读(172) 评论(0) 推荐(0)

leetcode228

摘要：1 class Solution: 2 def summaryRanges(self, nums: List[int]) -> List[str]: 3 n = len(nums) 4 if n == 0: 5 return [] 6 pre = nums[0] 7 res = [[nums[0], 阅读全文

posted @ 2020-04-04 20:34 Sempron2800+ 阅读(147) 评论(0) 推荐(0)

leetcode223

摘要：1 class Solution: 2 def computeArea(self, A: int, B: int, C: int, D: int, E: int, F: int, G: int, H: int) -> int: 3 x = 0 4 y = 0 5 6 if (A <= E): 7 i 阅读全文

posted @ 2020-04-04 10:14 Sempron2800+ 阅读(143) 评论(0) 推荐(0)

leetcode222

摘要：1 class Solution: 2 def __init__(self): 3 self.count = 0 4 5 def preOrder(self,node): 6 if node != None: 7 self.count += 1 8 self.preOrder(node.left) 阅读全文

posted @ 2020-04-04 09:45 Sempron2800+ 阅读(150) 评论(0) 推荐(0)

leetcode211

摘要：1 class TrieNode: 2 def __init__(self): 3 self.words = 0 4 self.edges = [None] * 26 5 6 7 class WordDictionary: 8 def __init__(self): 9 self.root = Tr 阅读全文

posted @ 2020-04-04 08:48 Sempron2800+ 阅读(156) 评论(0) 推荐(0)

leetcode209

摘要：1 class Solution { 2 public int minSubArrayLen(int s, int[] nums) { 3 int idx1=0,idx2=0; 4 int currSum=0; 5 int ret=Integer.MAX_VALUE; 6 while(idx2<nu 阅读全文

posted @ 2020-04-03 09:31 Sempron2800+ 阅读(185) 评论(0) 推荐(0)