topic
Given an undirected graph graph, return true when the graph is a bipartite graph.
If we can divide the set of nodes in a graph into two independent subsets A and B, and make the two nodes of each edge in the graph one from the set A and one from the set B, we call this graph binary picture.
The graph will be given as an adjacency list, and graph[i] represents all the nodes connected to node i in the graph. Each node is an integer between 0 and graph.length-1. There are no self-loops and parallel edges in this graph: i does not exist in graph[i], and there are no duplicate values in graph[i]. link
Ideas
Did not understand the question at first. . His adjacency list is a bit interesting, there are a total graph.lengthof nodes, and each subscript represents a node, that is [0,graph.length-1], graph[i]an array, which stores 节点ielements with edges between and. There is no idea. After reading the solution, the dyeing method can be solved. You can use dfs or bfs.
class Solution {
final static int RED = 1;
final static int GREEN = 2;
int [] colors;
/*bfs*/
public boolean isBipartite(int[][] graph) {
int num = graph.length;
colors = new int[num];
//从第一个节点开始
for(int i = 0; i < num; i++){
//遇到未染色的节点才开始bfs搜索
if(colors[i] == 0){
//未染色的都染为红色
colors[i] = RED;
Queue<Integer> queue = new LinkedList<>();
//从i开始bfs
queue.offer(i);
while(!queue.isEmpty()){
int curNode = queue.poll();
int color = colors[curNode];
//对其邻居进行染色
for(int neighbor: graph[curNode]){
//邻居未染色color为0，需要染上与curNode不同的颜色
if(colors[neighbor] == 0){
colors[neighbor] = color == RED ? GREEN : RED;
queue.offer(neighbor);
}//邻居染色了，若与curNode颜色相同直接返回
else if(colors[neighbor] == color){
return false;
}
}
}
}
}
return true;
}
}
class Solution {
final static int RED = 1;
final static int GREEN = 2;
int [] colors;
boolean valid = true;
/*dfs*/
public boolean isBipartite(int[][] graph) {
int num = graph.length;
colors = new int[num];
for(int i = 0; i < num; i++){
if(colors[i] == 0){
dfs(graph, i, RED);
}
}
return valid;
}
void dfs(int[][] graph, int curNode, int color){
if(!valid){
return;
}
colors[curNode] = color;
int neighborColor = color == RED ? GREEN : RED;
for(int neighbor: graph[curNode]){
if(colors[neighbor] == 0){
dfs(graph, neighbor, neighborColor);
}else if(colors[neighbor] != neighborColor){
valid = false;
return;
}
}
}
}