难度简单
示例 1:
输入:sentence = "i love eating burger", searchWord = "burg"
输出:4
解释:"burg" 是 "burger" 的前缀,而 "burger" 是句子中第 4 个单词。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution: def isPrefixOfWord(self, sentence: str, searchWord: str) -> int: l1 = len(sentence) i=0 j=0 index = 1 match=1 while i<l1: if sentence[i]==' ': index+=1 if match==1 and j>=len(searchWord): break match = 1 j=0 i+=1 #print(sentence[i],searchWord[j],match) elif sentence[i]==searchWord[j] and match: j+=1 i+=1 match =1 if j>=len(searchWord): break # print(sentence[i],searchWord[j],match) else: i+=1 match =0 if match ==0: return -1 return index
给你字符串 s 和整数 k 。
请返回字符串 s 中长度为 k 的单个子字符串中可能包含的最大元音字母数。
英文中的 元音字母 为(a, e, i, o, u)。
示例 1:
输入:s = "abciiidef", k = 3 输出:3 解释:子字符串 "iii" 包含 3 个元音字母。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution: def maxVowels(self, s: str, k: int) -> int: l1 = len(s) clist=['a','e','i','o','u'] t=0 max1 = [0]*l1 for i in range(k): if s[i] in ['a','e','i','o','u']: t+=1 max1[k-1] =t for i in range(k,l1): t = max1[i-1] if s[i-k] in clist: t-=1 if s[i] in clist: t+=1 max1[i] = t return max(max1)
第三题:
给你一棵二叉树,每个节点的值为 1 到 9 。我们称二叉树中的一条路径是 「伪回文」的,当它满足:路径经过的所有节点值的排列中,存在一个回文序列。
请你返回从根到叶子节点的所有路径中 伪回文 路径的数目。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/pseudo-palindromic-paths-in-a-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def pseudoPalindromicPaths (self, root: TreeNode) -> int: num =0 path=[] num = self.treepath(root,path,num) return num#-num//2 def treepath(self,node,path,num): import copy if node.left is None and node.right is None: path1 = copy.copy(path) path1.append(node.val) oddnum=0 rec={} # print(path1,node.val) for i in range(len(path1)): if path1[i] not in rec: rec[path1[i]]=0 rec[path1[i]]+=1 for c in rec: if rec[c]%2==1: oddnum+=1 if oddnum>1: return num # print(path,num) return num+1 # path.append(node.val) path1 = copy.copy(path) path1.append(node.val) if node.left is not None: num = self.treepath(node.left,path1,num) if node.right is not None: num = self.treepath(node.right,path1,num) return num