依旧是第 4 题不做（不会做，也不想花时间搞懂）。
第 1 题：检查单词是否为句中其他单词的前缀
https://leetcode-cn.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/
Java 代码：
public class Solution {
public int isPrefixOfWord(String sentence, String searchWord) {
String[] splits = sentence.split("\\s");
int len = splits.length;
for (int i = 0; i < len; i++) {
if (splits[i].indexOf(searchWord) == 0) {
return i + 1;
}
}
return -1;
}
public static void main(String[] args) {
String sentence = "hellohello hellohellohello";
String searchWord = "ell";
Solution solution = new Solution();
int res = solution.isPrefixOfWord(sentence, searchWord);
System.out.println(res);
}
}
第 2 题：定长子串中元音的最大数目
https://leetcode-cn.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/
知识点：固定长度的滑动窗口，其实要容易很多，注意一些细节。
Java 代码：
public class Solution {
public int maxVowels(String s, int k) {
char[] charArrayS = s.toCharArray();
int len = s.length();
int[] map = new int[128];
map['a'] = 1;
map['e'] = 1;
map['i'] = 1;
map['o'] = 1;
map['u'] = 1;
int count = 0;
int res = 0;
for (int i = 0; i < len; i++) {
if (map[charArrayS[i]] == 1) {
count++;
}
res = Math.max(res, count);
if (i >= k - 1) {
// 如果是元音，移除字符
if (map[charArrayS[i - (k - 1)]] == 1) {
count--;
}
}
}
return res;
}
public static void main(String[] args) {
// String s = "abciiidef";
// int k = 3;
// String s = "aeiou";
// int k = 2;
// String s = "leetcode";
// int k = 3;
String s = "rhythms";
int k = 4;
Solution solution = new Solution();
int res = solution.maxVowels(s, k);
System.out.println(res);
}
}
第 3 题：二叉树中的伪回文路径
相关知识点：
- 「力扣」第 113 题：路径总和 II；
- 位运算表示回文，这里涉及到状态压缩的知识点；
- 还涉及了回溯里整型状态的设计。
Java 代码：
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {
}
TreeNode(int val) {
this.val = val;
}
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
class Solution {
private int count = 0;
public int pseudoPalindromicPaths(TreeNode root) {
dfs(root, 0);
return count;
}
private void dfs(TreeNode node, int status) {
if (node == null) {
return;
}
status ^= (1 << node.val);
if (node.left == null && node.right == null) {
if (status == 0 || (status & (status - 1)) == 0) {
count++;
}
return;
}
if (node.left != null) {
dfs(node.left, status);
}
if (node.right != null) {
dfs(node.right, status);
}
}
}