263 lines
6.7 KiB
Java
263 lines
6.7 KiB
Java
<!-- GFM-TOC -->
|
||
* [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)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
# 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<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)](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<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)](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);
|
||
}
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
<img width="650px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/公众号海报1.png"></img>
|