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摘要: 题目描述: 方法一:递归 class Solution: def mirrorTree(self, root: TreeNode) -> TreeNode: if not root:return root.left,root.right = self.mirrorTree(root.right),s 阅读全文
posted @ 2020-04-21 19:41 oldby 阅读(102) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法:动态规划: class Solution: def numOfArrays(self, n: int, m: int, k: int) -> int: mod = 10 ** 9 + 7 dp = [[[0] * (k + 1) for _ in range(m + 1)] for 阅读全文
posted @ 2020-04-21 11:17 oldby 阅读(222) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 自己的提交: class Solution: def minNumberOfFrogs(self, croakOfFrogs: str) -> int: c = collections.Counter() ans = 0 st = "croak" for s in croakOfFrog 阅读全文
posted @ 2020-04-20 20:50 oldby 阅读(336) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 自己的提交: class Solution: def displayTable(self, orders: List[List[str]]) -> List[List[str]]: table = set() table_l = [] food = set() food_l = [] f 阅读全文
posted @ 2020-04-20 20:45 oldby 阅读(204) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 第一次提交: class Solution: def reformat(self, s: str) -> str: digit = [] alpha = [] for i in s: if i.isdigit(): digit.append(i) elif i.isalpha(): al 阅读全文
posted @ 2020-04-20 20:29 oldby 阅读(163) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: class Solution: def getMaxRepetitions(self, s1: str, n1: int, s2: str, n2: int) -> int: if n1 == 0: return 0 s1cnt, index, s2cnt = 0, 0, 0 阅读全文
posted @ 2020-04-20 13:09 oldby 阅读(116) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 解法:动态规划 class Solution: def numberOfArrays(self, s: str, k: int) -> int: n = len(s) dp = [0] * (n + 1) dp[0] = 1 mod = 10**9 +7 for i in range(1 阅读全文
posted @ 2020-04-19 12:57 oldby 阅读(254) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 提交: class Solution: def getHappyString(self, n: int, k: int) -> str: strlist = [] def helper(n, s): if len(s) == n: strlist.append(s) return for 阅读全文
posted @ 2020-04-19 10:29 oldby 阅读(376) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 解法:贪心,每次减去<k的最大斐波那契数 class Solution: def findMinFibonacciNumbers(self, k: int) -> int: fib = [1, 1] while fib[-1] <= k: fib.append(fib[-1] + fib 阅读全文
posted @ 2020-04-19 10:06 oldby 阅读(205) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 提交: class Solution: def minStartValue(self, nums: List[int]) -> int: res = 1 s = 0 for i in nums: s += i res = min(res,s) if res >= 0: return 1 阅读全文
posted @ 2020-04-19 09:50 oldby 阅读(108) 评论(0) 推荐(0) 编辑
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