随笔分类 - leetcode
摘要：题目描述： 方法：动态规划 完全背包问题 class Solution: def waysToChange(self, n: int) -> int: coins = [1,5,10,25] dp = [0] * (n+1) dp[0] = 1 for coin in coins: for i in
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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摘要：题目描述： 方法一：O(MN) O(M) class Solution: def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool: def equal(A,B): if not B:return True if not A or A.va
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摘要：方法一：迭代 class Solution { public ListNode reverseList(ListNode head) { ListNode pre = null; ListNode cur = head; ListNode tmp = null; while(cur!=null){
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摘要：题目描述： java：快慢指针： /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */
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摘要：题目描述： 方法一：dfs O(MN) O(MN) class Solution: def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]: m,n = len(matrix),len(matrix[0]) dist =
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摘要：方法一：栈 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solut
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摘要：题目描述： 方法一：逻辑判断法：O(N) class Solution: def isNumber(self, s: str) -> bool: s = s.strip() met_dot = met_e = met_digit = False for i, char in enumerate(s)
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摘要：题目描述： 方法： class Solution: def intersection(self, start1, end1, start2, end2): x1, y1, x2, y2, x3, y3, x4, y4 = *start1, *end1, *start2, *end2 det = la
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摘要：题目描述： 方法一： class Solution: def numOfWays(self, n: int) -> int: mod = int(1e9+7) s = ['ryr','ryg','rgr','rgy','yrg','ygr','ygy','yry','gry','grg','gyr'
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摘要：题目描述： 自己的提交： class Solution: def entityParser(self, text: str) -> str: text = text.replace(""","\"") text = text.replace("'","\'") text = te
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