資訊專欄INFORMATION COLUMN
Problem
Given an undirected graph, return true if and only if it is bipartite.
Recall that a graph is bipartite if we can split it"s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.
The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn"t contain any element twice.
Example 1:
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.
Note:
graph will have length in range [1, 100]. graph[i] will contain integers in range [0, graph.length - 1]. graph[i] will not contain i or duplicate values. The graph is undirected: if any element j is in graph[i], then i will be in graph[j].Solution DFS
class Solution {
public boolean isBipartite(int[][] graph) {
int[] colors = new int[graph.length];
for (int i = 0; i < graph.length; i++) {
if (colors[i] == 0 && !dfs(graph, colors, i, 1)) return false;
}
return true;
}
private boolean dfs(int[][] graph, int[] colors, int node, int color) {
if (colors[node] != 0) return colors[node] == color;
else colors[node] = color;
for (int neighbor: graph[node]) {
if (!dfs(graph, colors, neighbor, -color)) return false;
}
return true;
}
}
BFS
class Solution {
public boolean isBipartite(int[][] graph) {
int[] colors = new int[graph.length];
for (int i = 0; i < graph.length; i++) {
if (graph[i].length != 0 && colors[i] == 0) {
colors[i] = 1;
Queue queue = new LinkedList<>();
queue.offer(i);
while (!queue.isEmpty()) {
int node = queue.poll();
for (int neighbor: graph[node]) {
if (colors[neighbor] == 0) {
colors[neighbor] = -colors[node];
queue.offer(neighbor);
} else {
if (colors[neighbor] == colors[node]) return false;
}
}
}
}
}
return true;
}
}
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