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题目描述: 方法一:动态规划 O(mnlogmn) class Solution(object): def longestIncreasingPath(self, matrix): if not matrix or not matrix[0]: return 0 m, n = len(matrix) 阅读全文
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题目描述: 方法:二分法 O(N*log(summ - maxn)) class Solution: def splitArray(self, nums: List[int], m: int) -> int: n = len(nums) s = 0 for num in nums: s += num 阅读全文
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题目描述: 官方解答: 方法一:记忆化递归 O(N3) O(N2) class Solution: def maxCoins(self, nums: List[int]) -> int: n = len(nums) val = [1] + nums + [1] @lru_cache(None) de 阅读全文
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恢复内容开始 题目描述: 方法一:边统计边压缩: class Solution { public int numSubmat(int[][] mat) { int row = mat.length, col = mat[0].length, ans = 0; for (int i = 0; i < 阅读全文
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题目描述: 第一次提交:通过76/78 留坑 class Solution: def isBipartite(self, graph: List[List[int]]) -> bool: dic = {} l = [] for i in range(len(graph)): if graph[i] 阅读全文
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题目描述: 方法:动态规划 class Solution: def winnerSquareGame(self, n: int) -> bool: dp = [False, True, False] for x in range(3, n+1): dp.append(False) for y in 阅读全文
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题目描述: 提交: class Solution: def minDifference(self, nums: List[int]) -> int: if len(nums) <= 4: return 0 minm = min(nums) maxm = max(nums) l1 = heapq.nl 阅读全文
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题目描述: 提交:O(m+n) class Solution: def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]: nums2dic = collections.Counter(nums2) res = [] f 阅读全文
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题目描述: 方法一:动态规划 O(mn) O(mn) class Solution: def calculateMinimumHP(self, dungeon: List[List[int]]) -> int: rows = len(dungeon) cols = len(dungeon[0]) d 阅读全文
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题目描述: 提交:堆 class Solution: def minDifference(self, nums: List[int]) -> int: if len(nums) <= 4: return 0 minm = min(nums) maxm = max(nums) l1 = heapq.n 阅读全文