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)
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}. 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. 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
2, [[1,0]] return true 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。
public boolean canFinish(int numCourses, int[][] prerequisites) { List<Integer>[] 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<Integer>[] 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)
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。
public int[] findOrder(int numCourses, int[][] prerequisites) { List<Integer>[] 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<Integer> 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<Integer>[] graphic, int curNode, Stack<Integer> 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)
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。
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); } }