0. 前言
- 中文版地址：https://leetcode-cn.com/contest/weekly-contest-190/
- 英文版地址：https://leetcode.com/contest/weekly-contest-190/
1. 题解
1.1 5416. 检查单词是否为句中其他单词的前缀（1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence）
- 中文版题目描述：https://leetcode-cn.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/
- 英文版题目描述：https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/
- 思路：签到题，模拟
- 代码如下：
代码语言：javascript
复制class Solution {
public:
int isPrefixOfWord(string sentence, string searchWord) {
int len = sentence.length();
if (sentence[len-1] != ' ') {
sentence = " ";
len ;
}
int last = 0;
vector<string> words;
for (int i = 1 ; i< len ; i ) {
if (sentence[i] == ' ') {
words.push_back(sentence.substr(last, i-last));
last = i 1;
}
}
for (int i = 0 ; i < (int)words.size() ; i ) {
int lv = words[i].length(), ls = searchWord.length();
if (lv < ls) continue;
if (words[i].substr(0, ls) == searchWord) return i 1;
}
return -1;
}
};
1.2 5417. 定长子串中元音的最大数目（1456. Maximum Number of Vowels in a Substring of Given Length）
- 中文版题目描述：https://leetcode-cn.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/
- 英文版题目描述：https://leetcode.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/
- 思路：窗口统计，在一个大小为 k 的窗口内统计 'a'、'e'、'r'、'o'、'u' 个数
- 暴力代码如下：
代码语言：javascript
复制class Solution {
public:
int maxVowels(string s, int k) {
int len = (int)s.length(), cnt = 0, ans = 0;
for (int i = 0 ; i < k ; i ) {
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u') {
cnt ;
}
}
ans = cnt;
for (int i = k ; i < len ; i ) {
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u') {
cnt ;
}
if (s[i-k] == 'a' || s[i-k] == 'e' || s[i-k] == 'i' || s[i-k] == 'o' || s[i-k] == 'u') {
cnt--;
}
ans = max(ans, cnt);
}
return ans;
}
};
1.3 5418. 二叉树中的伪回文路径（1457. Pseudo-Palindromic Paths in a Binary Tree）
- 中文版题目描述：https://leetcode-cn.com/problems/pseudo-palindromic-paths-in-a-binary-tree/
- 英文版题目描述：https://leetcode.com/problems/pseudo-palindromic-paths-in-a-binary-tree/
- 思路：遍历二叉树，用一个数组统计每条从根节点到叶子节点路径 1-9 数字的个数
- 如果节点个数是奇数个，那么自身数量为奇数的数字个数只能为 1
- 如果节点个数是偶数个，那么自身数量为偶数的数字个数只能为 0
- 看某一个大神的题解，也可以通过 位运算实现，膜拜一下
- 代码如下：
代码语言：javascript
复制/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void dfs(TreeNode* root, vector<int>& cnt, int& ans) {
if (root->left == NULL && root->right == NULL) {
int sum = 0, odd = 0;
for (int i = 1 ; i <= 9 ; i ) {
if (cnt[i] % 2) odd ;
sum = cnt[i];
}
if (sum % 2 == odd) ans ;
return;
}
if (root->left) {
cnt[root->left->val] ;
dfs(root->left, cnt, ans);
cnt[root->left->val]--;
}
if (root->right) {
cnt[root->right->val] ;
dfs(root->right, cnt, ans);
cnt[root->right->val]--;
}
}
int pseudoPalindromicPaths (TreeNode* root) {
if (root == NULL) return 0;
vector<int> cnt = vector<int>(10, 0);
cnt[root->val] ;
int ans = 0;
dfs(root, cnt, ans);
cnt[root->val]--;
return ans;
}
};
1.4 5419. 两个子序列的最大点积（1458. Max Dot Product of Two Subsequences）
- 中文版题目描述：https://leetcode-cn.com/problems/max-dot-product-of-two-subsequences/
- 英文版题目描述：https://leetcode.com/problems/max-dot-product-of-two-subsequences/
- 思路：类似最长公共子序列，dp[i][j] 表示 nums1 前 i 个和 nums2 前 j 个的最大点积和，O(n^2) 复杂度
- 注意题中要求需要为非空子序列，因此初始值不应该为 0
- 一开始想复杂了，多加了一维表示子序列长度，结果一直超时，，想了半天恍然大明白
- 代码如下：
代码语言：javascript
复制class Solution {
public:
int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
int n1 = (int)nums1.size(), n2 = (int)nums2.size(), len = min(n1, n2);
vector<vector<int>> dp = vector<vector<int>>(n1 1, vector<int>(n2 1, -50000000));
int ans = INT_MIN;
for (int i = 0 ; i <= n1 ; i ) {
dp[i][0] = 0;
}
for (int i = 0 ; i <= n2 ; i ) {
dp[0][i] = 0;
}
for (int j = 1 ; j <= n1 ; j ) {
for (int k = 1 ; k <= n2 ; k ) {
if (j > 1 && k > 1) {
dp[j][k] = dp[j-1][k-1];
}
if (j > 1) {
dp[j][k] = max(dp[j][k], dp[j-1][k]);
}
if (k > 1) {
dp[j][k] = max(dp[j][k], dp[j][k-1]);
}
dp[j][k] = max(dp[j][k], max(dp[j-1][k-1] nums1[j-1] * nums2[k-1], nums1[j-1] * nums2[k-1]));
// cout << j << "<>" << k << "==" << dp[j][k] << " | " << ans << endl;
ans = max(ans, dp[j][k]);
}
}
return ans;
}
};
3. 参考文献
- https://leetcode-cn.com/problems/pseudo-palindromic-paths-in-a-binary-tree/solution/wei-yun-suan-jie-fa-by-dnanki/
- https://leetcode-cn.com/problems/max-dot-product-of-two-subsequences/solution/c-dong-tai-gui-hua-yi-dong-by-smilyt_/
