随笔分类 -  leetcode

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摘要:恢复内容开始 题目描述: class Solution: def shortestAlternatingPaths(self, n: int, red_edges, blue_edges): def function(n,r,b): result=[[float("inf")]*2 for _ in 阅读全文
posted @ 2019-07-24 09:58 oldby 阅读(258) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution(object): def numEquivDominoPairs(self, dominoes): """ :type dominoes: List[List[int]] :rtype: int """ f = {} ret = 0 for d i 阅读全文
posted @ 2019-07-21 15:21 oldby 阅读(170) 评论(0) 推荐(0)
摘要:题目描述: 第一次提交:BFS O(N) O(N) # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # 阅读全文
posted @ 2019-07-20 21:03 oldby 阅读(138) 评论(0) 推荐(0)
摘要:题目描述: 方法一:递归 # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = 阅读全文
posted @ 2019-07-20 20:41 oldby 阅读(232) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution(object): def rob(self, nums): """ :type nums: List[int] :rtype: int """ def dp(nums): pre = 0 cur = 0 for i in nums: temp = 阅读全文
posted @ 2019-07-20 19:32 oldby 阅读(170) 评论(0) 推荐(0)
摘要:题目描述: 方法一:O(N) O(N) class Solution(object): def rob(self, nums): """ :type nums: List[int] :rtype: int """ bp = [0] * (len(nums) + 2) for i in range(l 阅读全文
posted @ 2019-07-20 19:05 oldby 阅读(144) 评论(0) 推荐(0)
摘要:题目描述; 第一次提交: class Solution(object): def hammingWeight(self, n): """ :type n: int :rtype: int """ n = str(bin(n)) count = 0 for i in n: if i == '1': c 阅读全文
posted @ 2019-07-20 18:36 oldby 阅读(186) 评论(0) 推荐(0)
摘要:题目描述: 方法一:内置函数 class Solution: # @param n, an integer # @return an integer def reverseBits(self, n): return int(bin(n)[2:].zfill(32)[::-1],2) 方法二:位运算 阅读全文
posted @ 2019-07-20 17:38 oldby 阅读(137) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution: def rotate(self, nums: List[int], k: int) -> None: """ Do not return anything, modify nums in-place instead. """ for i in r 阅读全文
posted @ 2019-07-20 14:30 oldby 阅读(185) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution: def findRepeatedDnaSequences(self, s: str) -> List[str]: n = 10 d = {} for i in range(len(s)-9): if s[i:i+10] in d: d[s[i:i 阅读全文
posted @ 2019-07-20 13:51 oldby 阅读(262) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class LargerNumKey(str): def __lt__(x, y): return x+y > y+x class Solution: def largestNumber(self, nums): largest_num = ''.join(sorted(map 阅读全文
posted @ 2019-07-20 13:32 oldby 阅读(282) 评论(0) 推荐(0)
摘要:题目描述: 方法一: # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class 阅读全文
posted @ 2019-07-20 09:42 oldby 阅读(146) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution: def fractionToDecimal(self, numerator: int, denominator: int) -> str: if numerator == 0: return "0" res = [] if (numerator 阅读全文
posted @ 2019-07-18 14:46 oldby 阅读(188) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution: def compareVersion(self, version1: str, version2: str) -> int: v1, v2 = ([*map(int, v.split('.'))] for v in (version1, vers 阅读全文
posted @ 2019-07-18 13:22 oldby 阅读(248) 评论(0) 推荐(0)
摘要:题目描述: 二分法: class Solution: def findPeakElement(self, nums: List[int]) -> int: l, h = 0, len(nums) - 1 while l <= h: m = (l + h) // 2 if (not m or nums 阅读全文
posted @ 2019-07-18 11:38 oldby 阅读(131) 评论(0) 推荐(0)
摘要:题目描述: 方法一:辅助栈 class MinStack: def __init__(self): """ initialize your data structure here. """ self.stack = [] self.min_stack = [] def push(self, x: i 阅读全文
posted @ 2019-07-18 10:57 oldby 阅读(163) 评论(0) 推荐(0)
摘要:题目描述: 方法一: class Solution: def findMin(self, nums: List[int]) -> int: left, right = 0, len(nums) - 1 while left < right: mid = (left + right) // 2 if 阅读全文
posted @ 2019-07-17 20:52 oldby 阅读(163) 评论(0) 推荐(0)
摘要:题目描述: 第一次提交:O(N) class Solution: def findMin(self, nums: List[int]) -> int: if not nums: return l,r = 0,len(nums)-1 while l<r: if nums[l]>nums[r]: l + 阅读全文
posted @ 2019-07-17 20:42 oldby 阅读(220) 评论(0) 推荐(0)
摘要:题目描述: 方法一:动态规划 class Solution: def maxProduct(self, nums: List[int]) -> int: ret,up,down=nums[0],nums[0],nums[0] for n in nums[1:]: if n>=0: up,down=m 阅读全文
posted @ 2019-07-17 20:22 oldby 阅读(104) 评论(0) 推荐(0)
摘要:题目描述: 方法一: return " ".join(s.split()[::-1]) 方法二: class Solution: def reverseWords(self, s: str) -> str: res = re.findall('\S+',s) return " ".join(res[ 阅读全文
posted @ 2019-07-17 19:47 oldby 阅读(224) 评论(0) 推荐(0)

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