2019-04-25 18:24:51 +08:00
<!-- GFM - TOC -->
* [二分图 ](#二分图 )
2019-05-14 22:56:30 +08:00
* [1. 判断是否为二分图 ](#1-判断是否为二分图 )
2019-04-25 18:24:51 +08:00
* [拓扑排序 ](#拓扑排序 )
2019-05-14 22:56:30 +08:00
* [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 )
2019-04-25 18:24:51 +08:00
* [并查集 ](#并查集 )
2019-05-14 22:56:30 +08:00
* [1. 冗余连接 ](#1-冗余连接 )
2019-04-25 18:24:51 +08:00
<!-- 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。
2019-05-14 22:56:30 +08:00
## 1. 判断是否为二分图
2019-04-25 18:24:51 +08:00
[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。
2019-05-14 22:56:30 +08:00
## 1. 课程安排的合法性
2019-04-25 18:24:51 +08:00
[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;
}
```
2019-05-14 22:56:30 +08:00
## 2. 课程安排的顺序
2019-04-25 18:24:51 +08:00
[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： ： ，
```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。
2019-05-14 22:56:30 +08:00
## 1. 冗余连接
2019-04-25 18:24:51 +08:00
[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);
}
}
```
2019-06-12 12:20:21 +08:00
# 和我交流
2019-06-10 11:23:18 +08:00
2019-06-12 12:20:21 +08:00
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。
2019-06-10 11:23:18 +08:00
2019-06-12 12:26:39 +08:00
< div align = "center" > < img width = "580px" src = "https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/公众号海报2.png" > < / img > < / div >
