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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(129) 评论(0) 推荐(0) 编辑
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(178) 评论(0) 推荐(0) 编辑
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(253) 评论(0) 推荐(0) 编辑
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(278) 评论(0) 推荐(0) 编辑
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(133) 评论(0) 推荐(0) 编辑
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(180) 评论(0) 推荐(0) 编辑
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(242) 评论(0) 推荐(0) 编辑
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(126) 评论(0) 推荐(0) 编辑
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(158) 评论(0) 推荐(0) 编辑
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(158) 评论(0) 推荐(0) 编辑
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