684:冗余连接
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src/main/java/leetcode/editor/cn/RedundantConnection.java
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src/main/java/leetcode/editor/cn/RedundantConnection.java
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//<p>树可以看成是一个连通且 <strong>无环 </strong>的 <strong>无向 </strong>图。</p>
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//
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//<p>给定往一棵 <code>n</code> 个节点 (节点值 <code>1～n</code>) 的树中添加一条边后的图。添加的边的两个顶点包含在 <code>1</code> 到 <code>n</code> 中间，且这条附加的边不属于树中已存在的边。图的信息记录于长度为 <code>n</code> 的二维数组 <code>edges</code> ，<code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示图中在 <code>ai</code> 和 <code>bi</code> 之间存在一条边。</p>
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//
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//<p>请找出一条可以删去的边，删除后可使得剩余部分是一个有着 <code>n</code> 个节点的树。如果有多个答案，则返回数组 <code>edges</code> 中最后出现的边。</p>
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//
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//<p> </p>
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//
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//<p><strong>示例 1：</strong></p>
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//
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//<p><img alt="" src="https://pic.leetcode-cn.com/1626676174-hOEVUL-image.png" style="width: 152px; " /></p>
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//
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//<pre>
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//<strong>输入:</strong> edges = [[1,2], [1,3], [2,3]]
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//<strong>输出:</strong> [2,3]
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//</pre>
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//
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//<p><strong>示例 2：</strong></p>
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//
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//<p><img alt="" src="https://pic.leetcode-cn.com/1626676179-kGxcmu-image.png" style="width: 250px; " /></p>
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//
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//<pre>
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//<strong>输入:</strong> edges = [[1,2], [2,3], [3,4], [1,4], [1,5]]
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//<strong>输出:</strong> [1,4]
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//</pre>
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//
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//<p> </p>
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//
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//<p><strong>提示:</strong></p>
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//
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//<ul>
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// <li><code>n == edges.length</code></li>
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// <li><code>3 <= n <= 1000</code></li>
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// <li><code>edges[i].length == 2</code></li>
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// <li><code>1 <= ai < bi <= edges.length</code></li>
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// <li><code>ai != bi</code></li>
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// <li><code>edges</code> 中无重复元素</li>
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// <li>给定的图是连通的 </li>
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//</ul>
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//
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//<div><div>Related Topics</div><div><li>深度优先搜索</li><li>广度优先搜索</li><li>并查集</li><li>图</li></div></div><br><div><li>👍 523</li><li>👎 0</li></div>
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package leetcode.editor.cn;
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import java.util.*;
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// 684:冗余连接
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public class RedundantConnection {
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public static void main(String[] args) {
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Solution solution = new RedundantConnection().new Solution();
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// TO TEST
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}
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int[] findRedundantConnection(int[][] edges) {
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List<List<Integer>> list = new ArrayList<>();
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for (int[] edge : edges) {
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if (list.size() == 0) {
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list.add(new ArrayList<>(Arrays.asList(edge[0], edge[1])));
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} else {
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int index0 = -1;
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int index1 = -1;
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for (int i = list.size() - 1; i >= 0; i--) {
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if (list.get(i).contains(edge[0])) {
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index0 = i;
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if (index1 > -1) {
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break;
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}
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}
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if (list.get(i).contains(edge[1])) {
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index1 = i;
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if (index0 > -1) {
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break;
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}
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}
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}
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if (index0 > -1 && index0 == index1) {
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return edge;
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}
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Set<Integer> tmp = new HashSet<>();
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if (index0 > -1) {
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tmp.addAll(list.get(index0));
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} else {
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tmp.add(edge[0]);
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}
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if (index1 > -1) {
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tmp.addAll(list.get(index1));
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} else {
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tmp.add(edge[1]);
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}
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if (index0 > -1 && index1 > -1) {
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if (index0 > index1) {
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list.remove(index0);
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list.remove(index1);
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} else {
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list.remove(index1);
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list.remove(index0);
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}
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} else if (index0 > -1) {
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list.remove(index0);
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} else if (index1 > -1) {
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list.remove(index1);
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}
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list.add(new ArrayList<>(tmp));
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}
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}
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return null;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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@ -0,0 +1,41 @@
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<p>树可以看成是一个连通且 <strong>无环 </strong>的 <strong>无向 </strong>图。</p>
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<p>给定往一棵 <code>n</code> 个节点 (节点值 <code>1～n</code>) 的树中添加一条边后的图。添加的边的两个顶点包含在 <code>1</code> 到 <code>n</code> 中间，且这条附加的边不属于树中已存在的边。图的信息记录于长度为 <code>n</code> 的二维数组 <code>edges</code> ，<code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示图中在 <code>ai</code> 和 <code>bi</code> 之间存在一条边。</p>
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<p>请找出一条可以删去的边，删除后可使得剩余部分是一个有着 <code>n</code> 个节点的树。如果有多个答案，则返回数组 <code>edges</code> 中最后出现的边。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<p><img alt="" src="https://pic.leetcode-cn.com/1626676174-hOEVUL-image.png" style="width: 152px; " /></p>
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<pre>
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<strong>输入:</strong> edges = [[1,2], [1,3], [2,3]]
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<strong>输出:</strong> [2,3]
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</pre>
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<p><strong>示例 2：</strong></p>
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<p><img alt="" src="https://pic.leetcode-cn.com/1626676179-kGxcmu-image.png" style="width: 250px; " /></p>
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<pre>
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<strong>输入:</strong> edges = [[1,2], [2,3], [3,4], [1,4], [1,5]]
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<strong>输出:</strong> [1,4]
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</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>n == edges.length</code></li>
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<li><code>3 <= n <= 1000</code></li>
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<li><code>edges[i].length == 2</code></li>
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<li><code>1 <= ai < bi <= edges.length</code></li>
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<li><code>ai != bi</code></li>
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<li><code>edges</code> 中无重复元素</li>
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<li>给定的图是连通的 </li>
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</ul>
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<div><div>Related Topics</div><div><li>深度优先搜索</li><li>广度优先搜索</li><li>并查集</li><li>图</li></div></div><br><div><li>👍 523</li><li>👎 0</li></div>
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