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摘要: 题目描述: 方法一:迭代 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: res,stack = [],[] while True: while root: stack.append(root) roo 阅读全文
posted @ 2019-07-13 18:33 oldby 阅读(154) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一:暴力法 class Solution: def restoreIpAddresses(self, s: str) -> List[str]: n = len(s) res = [] # 判读是否满足ip的条件 def helper(tmp): if not tmp or (tmp 阅读全文
posted @ 2019-07-13 17:12 oldby 阅读(301) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: class Solution: def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode: dummy = ListNode(0) dummy.next = head pre = dummy for 阅读全文
posted @ 2019-07-13 15:57 oldby 阅读(85) 评论(0) 推荐(0) 编辑
摘要: 恢复内容开始 题目描述: 方法一:回溯(超时) class Solution: def numDecodings(self, s: str) -> int: res = 0 def backtrack(s): nonlocal res if not s: res += 1 return if int 阅读全文
posted @ 2019-07-13 14:50 oldby 阅读(228) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: class Solution: def grayCode(self, n: int) -> List[int]: gray = [0] for i in range(n): add = 2**i for j in range(len(gray)-1,-1,-1): gray.a 阅读全文
posted @ 2019-07-13 13:59 oldby 阅读(135) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def parti 阅读全文
posted @ 2019-07-13 12:51 oldby 阅读(130) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 33题 方法一: class Solution: def search(self, nums: List[int], target: int) -> bool: l, r = 0, len(nums) - 1 while l <= r: m = (l+r) // 2 if target 阅读全文
posted @ 2019-07-13 11:54 oldby 阅读(416) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 第一次提交: class Solution: def combine(self, n: int, k: int) -> List[List[int]]: res = [] def backtrack(i,temp_list): if len(temp_list)==k: res.appe 阅读全文
posted @ 2019-07-13 10:56 oldby 阅读(232) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: class Solution: def sortColors(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ p0 = curr = 0 阅读全文
posted @ 2019-07-13 10:34 oldby 阅读(164) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 第一次提交: class Solution: def removeDuplicates(self, nums: List[int]) -> int: nums.reverse() for i in range(len(nums)-3,-1,-1): if nums[i]==nums[i+ 阅读全文
posted @ 2019-07-13 10:16 oldby 阅读(170) 评论(0) 推荐(0) 编辑
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