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Commit ce2a9c2

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feat: add weekly contest 410 (#3401)
1 parent dc4a9d5 commit ce2a9c2

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lines changed

37 files changed

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‎solution/1000-1099/1035.Uncrossed Lines/README.md‎

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<pre>
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<strong>输入:</strong>nums1 = <span id="example-input-1-1">[1,4,2]</span>, nums2 = <span id="example-input-1-2">[1,2,4]</span>
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<strong>输出:</strong><span id="example-output-1">2</span>
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<strong>解释:</strong>可以画出两条不交叉的线,如上图所示。
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<strong>解释:</strong>可以画出两条不交叉的线,如上图所示。
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但无法画出第三条不相交的直线,因为从 nums1[1]=4 到 nums2[2]=4 的直线将与从 nums1[2]=2 到 nums2[1]=2 的直线相交。
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</pre>
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‎solution/3200-3299/3248.Snake in Matrix/README.md‎

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‎solution/3200-3299/3248.Snake in Matrix/README_EN.md‎

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class Solution {
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public:
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int finalPositionOfSnake(int n, vector<string>& commands) {
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int x = 0, y = 0;
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for (const auto& c : commands) {
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switch (c[0]) {
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case 'U': x--; break;
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case 'D': x++; break;
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case 'L': y--; break;
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case 'R': y++; break;
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}
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}
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return x * n + y;
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}
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};
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func finalPositionOfSnake(n int, commands []string) int {
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x, y := 0, 0
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for _, c := range commands {
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switch c[0] {
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case 'U':
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x--
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case 'D':
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x++
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case 'L':
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y--
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case 'R':
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y++
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}
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}
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return x*n + y
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}
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class Solution {
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public int finalPositionOfSnake(int n, List<String> commands) {
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int x = 0, y = 0;
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for (var c : commands) {
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switch (c.charAt(0)) {
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case 'U' -> x--;
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case 'D' -> x++;
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case 'L' -> y--;
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case 'R' -> y++;
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}
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}
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return x * n + y;
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}
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}
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class Solution:
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def finalPositionOfSnake(self, n: int, commands: List[str]) -> int:
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x = y = 0
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for c in commands:
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match c[0]:
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case "U":
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x -= 1
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case "D":
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x += 1
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case "L":
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y -= 1
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case "R":
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y += 1
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return x * n + y
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function finalPositionOfSnake(n: number, commands: string[]): number {
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let [x, y] = [0, 0];
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for (const c of commands) {
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c[0] === 'U' && x--;
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c[0] === 'D' && x++;
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c[0] === 'L' && y--;
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c[0] === 'R' && y++;
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}
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return x * n + y;
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}
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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3249.Count%20the%20Number%20of%20Good%20Nodes/README.md
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---
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<!-- problem:start -->
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# [3249. 统计好节点的数目](https://leetcode.cn/problems/count-the-number-of-good-nodes)
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[English Version](/solution/3200-3299/3249.Count%20the%20Number%20of%20Good%20Nodes/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>现有一棵 <strong>无向</strong> 树,树中包含 <code>n</code> 个节点,按从 <code>0</code> 到 <code>n - 1</code> 标记。树的根节点是节点 <code>0</code> 。给你一个长度为 <code>n - 1</code> 的二维整数数组 <code>edges</code>,其中 <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示树中节点 <code>a<sub>i</sub></code> 与节点 <code>b<sub>i</sub></code> 之间存在一条边。</p>
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<p>如果一个节点的所有子节点为根的&nbsp;<span data-keyword="subtree">子树</span>&nbsp;包含的节点数相同,则认为该节点是一个 <strong>好节点</strong>。</p>
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<p>返回给定树中<strong> 好节点 </strong>的数量。</p>
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<p><strong>子树</strong>&nbsp;指的是一个节点以及它所有后代节点构成的一棵树。</p>
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<p>&nbsp;</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong><span class="example-io">edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]</span></p>
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<p><strong>输出:</strong><span class="example-io">7</span></p>
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<p><strong>说明:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3249.Count%20the%20Number%20of%20Good%20Nodes/images/tree1.png" style="width: 360px; height: 158px;" />
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<p>树的所有节点都是好节点。</p>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong><span class="example-io">edges = [[0,1],[1,2],[2,3],[3,4],[0,5],[1,6],[2,7],[3,8]]</span></p>
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<p><strong>输出:</strong><span class="example-io">6</span></p>
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<p><strong>说明:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3249.Count%20the%20Number%20of%20Good%20Nodes/images/screenshot-2024年06月03日-193552.png" style="width: 360px; height: 303px;" />
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<p>树中有 6 个好节点。上图中已将这些节点着色。</p>
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</div>
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<p><strong class="example">示例 3:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>edges = [[0,1],[1,2],[1,3],[1,4],[0,5],[5,6],[6,7],[7,8],[0,9],[9,10],[9,12],[10,11]]</span></p>
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<p><span class="example-io"><b>输出:</b>12</span></p>
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<p><strong>解释:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3249.Count%20the%20Number%20of%20Good%20Nodes/images/rob.jpg" style="width: 450px; height: 277px;" />
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<p>除了节点 9 以外其他所有节点都是好节点。</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>edges.length == n - 1</code></li>
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<li><code>edges[i].length == 2</code></li>
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<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
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<li>输入确保 <code>edges</code> 总表示一棵有效的树。</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:DFS
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我们先根据题目给定的边 $\textit{edges}$ 构建出树的邻接表 $\textit{g},ドル其中 $\textit{g}[a]$ 表示节点 $a$ 的所有邻居节点。
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然后,我们设计一个函数 $\textit{dfs}(a, \textit{fa}),ドル表示计算以节点 $a$ 为根的子树中的节点数,并累计好节点的数量。其中 $\textit{fa}$ 表示节点 $a$ 的父节点。
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函数 $\textit{dfs}(a, \textit{fa})$ 的执行过程如下:
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1. 初始化变量 $\textit{pre} = -1,ドル $\textit{cnt} = 1,ドル $\textit{ok} = 1,ドル分别表示节点 $a$ 的某个子树的节点数、节点 $a$ 的所有子树的节点数、以及节点 $a$ 是否为好节点。
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2. 遍历节点 $a$ 的所有邻居节点 $b,ドル如果 $b$ 不等于 $\textit{fa},ドル则递归调用 $\textit{dfs}(b, a),ドル返回值为 $\textit{cur},ドル并累加到 $\textit{cnt}$ 中。如果 $\textit{pre} < 0,ドル则将 $\textit{cur}$ 赋值给 $\textit{pre}$;否则,如果 $\textit{pre}$ 不等于 $\textit{cur},ドル说明节点 $a$ 的不同子树的节点数不同,将 $\textit{ok}$ 置为 0ドル$。
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3. 最后,累加 $\textit{ok}$ 到答案中,并返回 $\textit{cnt}$。
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在主函数中,我们调用 $\textit{dfs}(0, -1),ドル最后返回答案。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 表示节点的数量。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def countGoodNodes(self, edges: List[List[int]]) -> int:
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def dfs(a: int, fa: int) -> int:
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pre = -1
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cnt = ok = 1
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for b in g[a]:
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if b != fa:
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cur = dfs(b, a)
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cnt += cur
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if pre < 0:
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pre = cur
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elif pre != cur:
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ok = 0
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nonlocal ans
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ans += ok
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return cnt
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g = defaultdict(list)
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for a, b in edges:
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g[a].append(b)
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g[b].append(a)
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ans = 0
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dfs(0, -1)
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return ans
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```
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#### Java
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```java
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class Solution {
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private int ans;
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private List<Integer>[] g;
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public int countGoodNodes(int[][] edges) {
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int n = edges.length + 1;
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g = new List[n];
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Arrays.setAll(g, k -> new ArrayList<>());
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for (var e : edges) {
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int a = e[0], b = e[1];
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g[a].add(b);
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g[b].add(a);
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}
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dfs(0, -1);
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return ans;
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}
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private int dfs(int a, int fa) {
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int pre = -1, cnt = 1, ok = 1;
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for (int b : g[a]) {
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if (b != fa) {
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int cur = dfs(b, a);
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cnt += cur;
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if (pre < 0) {
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pre = cur;
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} else if (pre != cur) {
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ok = 0;
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}
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}
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}
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ans += ok;
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return cnt;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int countGoodNodes(vector<vector<int>>& edges) {
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int n = edges.size() + 1;
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vector<int> g[n];
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for (const auto& e : edges) {
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int a = e[0], b = e[1];
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g[a].push_back(b);
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g[b].push_back(a);
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}
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int ans = 0;
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auto dfs = [&](auto&& dfs, int a, int fa) -> int {
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int pre = -1, cnt = 1, ok = 1;
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for (int b : g[a]) {
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if (b != fa) {
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int cur = dfs(dfs, b, a);
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cnt += cur;
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if (pre < 0) {
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pre = cur;
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} else if (pre != cur) {
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ok = 0;
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}
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}
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}
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ans += ok;
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return cnt;
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};
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dfs(dfs, 0, -1);
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return ans;
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}
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};
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```
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#### Go
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```go
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func countGoodNodes(edges [][]int) (ans int) {
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n := len(edges) + 1
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g := make([][]int, n)
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for _, e := range edges {
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a, b := e[0], e[1]
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g[a] = append(g[a], b)
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g[b] = append(g[b], a)
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}
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var dfs func(int, int) int
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dfs = func(a, fa int) int {
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pre, cnt, ok := -1, 1, 1
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for _, b := range g[a] {
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if b != fa {
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cur := dfs(b, a)
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cnt += cur
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if pre < 0 {
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pre = cur
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} else if pre != cur {
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ok = 0
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}
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}
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}
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ans += ok
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return cnt
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}
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dfs(0, -1)
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return
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}
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```
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#### TypeScript
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```ts
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function countGoodNodes(edges: number[][]): number {
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const n = edges.length + 1;
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const g: number[][] = Array.from({ length: n }, () => []);
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for (const [a, b] of edges) {
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g[a].push(b);
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g[b].push(a);
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}
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let ans = 0;
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const dfs = (a: number, fa: number): number => {
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let [pre, cnt, ok] = [-1, 1, 1];
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for (const b of g[a]) {
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if (b !== fa) {
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const cur = dfs(b, a);
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cnt += cur;
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if (pre < 0) {
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pre = cur;
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} else if (pre !== cur) {
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ok = 0;
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}
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}
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}
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ans += ok;
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return cnt;
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};
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dfs(0, -1);
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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