* [二分图](#二分图) * [1. 判断是否为二分图](#1-判断是否为二分图) * [拓扑排序](#拓扑排序) * [1. 课程安排的合法性](#1-课程安排的合法性) * [2. 课程安排的顺序](#2-课程安排的顺序) * [并查集](#并查集) * [1. 冗余连接](#1-冗余连接) # 二分图 如果可以用两种颜色对图中的节点进行着色，并且保证相邻的节点颜色不同，那么这个图就是二分图。 ## 1. 判断是否为二分图 [785. Is Graph Bipartite? (Medium)](https://leetcode.com/problems/is-graph-bipartite/description/) ```html Input: [[1,3], [0,2], [1,3], [0,2]] Output: true Explanation: The graph looks like this: 0----1 | | | | 3----2 We can divide the vertices into two groups: {0, 2} and {1, 3}. ``` ```html Example 2: Input: [[1,2,3], [0,2], [0,1,3], [0,2]] Output: false Explanation: The graph looks like this: 0----1 | \ | | \ | 3----2 We cannot find a way to divide the set of nodes into two independent subsets. ``` ```java public boolean isBipartite(int[][] graph) { int[] colors = new int[graph.length]; Arrays.fill(colors, -1); for (int i = 0; i < graph.length; i++) { // 处理图不是连通的情况 if (colors[i] == -1 && !isBipartite(i, 0, colors, graph)) { return false; } } return true; } private boolean isBipartite(int curNode, int curColor, int[] colors, int[][] graph) { if (colors[curNode] != -1) { return colors[curNode] == curColor; } colors[curNode] = curColor; for (int nextNode : graph[curNode]) { if (!isBipartite(nextNode, 1 - curColor, colors, graph)) { return false; } } return true; } ``` # 拓扑排序 常用于在具有先序关系的任务规划中。 ## 1. 课程安排的合法性 [207. Course Schedule (Medium)](https://leetcode.com/problems/course-schedule/description/) ```html 2, [[1,0]] return true ``` ```html 2, [[1,0],[0,1]] return false ``` 题目描述：一个课程可能会先修课程，判断给定的先修课程规定是否合法。 本题不需要使用拓扑排序，只需要检测有向图是否存在环即可。 ```java public boolean canFinish(int numCourses, int[][] prerequisites) { List[] graphic = new List[numCourses]; for (int i = 0; i < numCourses; i++) { graphic[i] = new ArrayList<>(); } for (int[] pre : prerequisites) { graphic[pre[0]].add(pre[1]); } boolean[] globalMarked = new boolean[numCourses]; boolean[] localMarked = new boolean[numCourses]; for (int i = 0; i < numCourses; i++) { if (hasCycle(globalMarked, localMarked, graphic, i)) { return false; } } return true; } private boolean hasCycle(boolean[] globalMarked, boolean[] localMarked, List[] graphic, int curNode) { if (localMarked[curNode]) { return true; } if (globalMarked[curNode]) { return false; } globalMarked[curNode] = true; localMarked[curNode] = true; for (int nextNode : graphic[curNode]) { if (hasCycle(globalMarked, localMarked, graphic, nextNode)) { return true; } } localMarked[curNode] = false; return false; } ``` ## 2. 课程安排的顺序 [210. Course Schedule II (Medium)](https://leetcode.com/problems/course-schedule-ii/description/) ```html 4, [[1,0],[2,0],[3,1],[3,2]] There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3]. ``` 使用 DFS 来实现拓扑排序，使用一个栈存储后序遍历结果，这个栈的逆序结果就是拓扑排序结果。 证明：对于任何先序关系：v->w，后序遍历结果可以保证 w 先进入栈中，因此栈的逆序结果中 v 会在 w 之前。 ```java public int[] findOrder(int numCourses, int[][] prerequisites) { List[] graphic = new List[numCourses]; for (int i = 0; i < numCourses; i++) { graphic[i] = new ArrayList<>(); } for (int[] pre : prerequisites) { graphic[pre[0]].add(pre[1]); } Stack postOrder = new Stack<>(); boolean[] globalMarked = new boolean[numCourses]; boolean[] localMarked = new boolean[numCourses]; for (int i = 0; i < numCourses; i++) { if (hasCycle(globalMarked, localMarked, graphic, i, postOrder)) { return new int[0]; } } int[] orders = new int[numCourses]; for (int i = numCourses - 1; i >= 0; i--) { orders[i] = postOrder.pop(); } return orders; } private boolean hasCycle(boolean[] globalMarked, boolean[] localMarked, List[] graphic, int curNode, Stack postOrder) { if (localMarked[curNode]) { return true; } if (globalMarked[curNode]) { return false; } globalMarked[curNode] = true; localMarked[curNode] = true; for (int nextNode : graphic[curNode]) { if (hasCycle(globalMarked, localMarked, graphic, nextNode, postOrder)) { return true; } } localMarked[curNode] = false; postOrder.push(curNode); return false; } ``` # 并查集 并查集可以动态地连通两个点，并且可以非常快速地判断两个点是否连通。 ## 1. 冗余连接 [684. Redundant Connection (Medium)](https://leetcode.com/problems/redundant-connection/description/) ```html Input: [[1,2], [1,3], [2,3]] Output: [2,3] Explanation: The given undirected graph will be like this: 1 / \ 2 - 3 ``` 题目描述：有一系列的边连成的图，找出一条边，移除它之后该图能够成为一棵树。 ```java public int[] findRedundantConnection(int[][] edges) { int N = edges.length; UF uf = new UF(N); for (int[] e : edges) { int u = e[0], v = e[1]; if (uf.connect(u, v)) { return e; } uf.union(u, v); } return new int[]{-1, -1}; } private class UF { private int[] id; UF(int N) { id = new int[N + 1]; for (int i = 0; i < id.length; i++) { id[i] = i; } } void union(int u, int v) { int uID = find(u); int vID = find(v); if (uID == vID) { return; } for (int i = 0; i < id.length; i++) { if (id[i] == uID) { id[i] = vID; } } } int find(int p) { return id[p]; } boolean connect(int u, int v) { return find(u) == find(v); } } ``` # 和我交流 如果你想和我交流，可以在我的微信公众号后台留言。另外，公众号提供了该项目的离线阅读版本，后台回复 "下载" 即可领取。也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。