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方法一： 二分法 O(nlogn) O(1) class Solution { public int findDuplicate(int[] nums) { int len = nums.length; int left = 1; int right = len - 1; while(left < 阅读全文
posted @ 2020-05-26 23:03
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题目描述： 提交： class Solution: def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int: n1,n2 = len(nums1),len(nums2) dp = [[0] * n2 for _ in ra 阅读全文
posted @ 2020-05-25 10:54
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题目描述： 提交： class Solution: def pseudoPalindromicPaths (self, root: TreeNode) -> int: def helper(l,node): left,right = 0,0 if node.left == None and node 阅读全文
posted @ 2020-05-25 10:47
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提交： class Solution: def maxVowels(self, s: str, k: int) -> int: n = len(s) dp = [0] * (n + 1) for i in range(n): if s[i] in "aeiou": dp[i + 1] = dp[i] 阅读全文
posted @ 2020-05-25 10:30
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题目描述： 提交： class Solution: def isPrefixOfWord(self, sentence: str, searchWord: str) -> int: n = len(searchWord) sentence_l = sentence.split(" ") for i 阅读全文
posted @ 2020-05-25 10:20
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题目描述： 方法一：动态规划 O(N) class Solution: def translateNum(self, num: int) -> int: str_num = str(num) n = len(str_num) dp = [1 for _ in range(n + 1)] for i 阅读全文
posted @ 2020-05-21 19:50
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题目描述： 方法：前缀和+状态压缩 O(N) class Solution: def findTheLongestSubstring(self, s: str) -> int: bit_mask = eval('0b00000') state_first_idx = {eval('0b00000') 阅读全文
posted @ 2020-05-20 20:08
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方法：自定义排序：O(logn) class Solution { public String minNumber(int[] nums) { String[] strs = new String[nums.length]; for(int i = 0; i < nums.length; i++) 阅读全文
posted @ 2020-05-19 21:16
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提交：O(N) O(1) class Solution: def validPalindrome(self, s: str) -> bool: def helper(s): if s == s[::-1]: return True return False l,r = 0,len(s) - 1 wh 阅读全文
posted @ 2020-05-19 20:44
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方法一： 来自讨论区 https://leetcode-cn.com/circle/discuss/Z2oiVE/ class Solution: def numPoints(self, p: List[List[int]], r: int) -> int: eps = 1e-8 def dist( 阅读全文
posted @ 2020-05-19 20:26
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