1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence 链接到标题
切分句子然后遍历判断是否为前缀，需要返回索引 + 1。
class Solution:
def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
for idx, w in enumerate(sentence.split(' ')):
if w.startswith(searchWord):
return idx + 1
else:
return -1
1456. Maximum Number of Vowels in a Substring of Given Length 链接到标题
滑动窗口，使用字典统计每个元音出现的次数，每次更新最大值。
class Solution:
def maxVowels(self, s: str, k: int) -> int:
d = {'a': 0, 'e': 0, 'i': 0, 'o': 0, 'u': 0}
for c in s[:k]:
if c in d:
d[c] += 1
res = sum(d.values())
for i in range(k, len(s)):
if s[i-k] in d:
d[s[i-k]] -= 1
if s[i] in d:
d[s[i]] += 1
res = max(res, sum(d.values()))
return res
1200. Minimum Absolute Difference 链接到标题
遍历列表，记录前一个数值，每次计算绝对差值，更新最小值，如果差值等于最小值，则将其加入到结果中。
class Solution:
def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
arr.sort()
m = float('inf')
out = []
for i in range(1, len(arr)):
prev = arr[i - 1]
curr = abs(prev - arr[i])
if curr < m:
out = [[prev, arr[i]]]
m = curr
elif curr == m: out.append([prev, arr[i]])
return out
1457. Pseudo-Palindromic Paths in a Binary Tree 链接到标题
DFS 遍历二叉树，使用 set 记录已有数值，如果该数值已经存在，则删掉，如果不存在，则加入set。判断回文字符的依据是set 中剩余字母长度是否小于等于1。
class Solution:
def pseudoPalindromicPaths (self, root: TreeNode) -> int:
s = set()
def dfs(root):
if not root: return 0
if root.val in s:
s.discard(root.val)
else:
s.add(root.val)
res = dfs(root.left) + dfs(root.right)
if not root.left and not root.right:
res += len(s)<=1
if root.val in s:
s.discard(root.val)
else:
s.add(root.val)
return res
return dfs(root)