785. Is Graph Bipartite? ⭐
Medium
There is an undirected graph with
n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:- There are no self-edges (
graph[u]does not containu). - There are no parallel edges (
graph[u]does not contain duplicate values). - If
vis ingraph[u], thenuis ingraph[v](the graph is undirected). - The graph may not be connected, meaning there may be two nodes
uandvsuch that there is no path between them.
A graph is bipartite if the nodes can be partitioned into two independent sets
A and B such that every edge in the graph connects a node in set A and a node in set B.Return
true if and only if it is bipartite.Example 1:
Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.
Example 2:
Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.
Constraints:
graph.length == n1 <= n <= 1000 <= graph[u].length < n0 <= graph[u][i] <= n - 1graph[u]does not containu.- All the values of
graph[u]are unique. - If
graph[u]containsv, thengraph[v]containsu.
func isBipartite(graph [][]int) bool {
// 著色，相鄰的點應該不同色
arr := make([]int, len(graph))
var dfs func(int, int) bool // 用來遍歷連通的點進行著色
dfs = func(n, color int) bool {
arr[n] = color
for _, node := range graph[n] {
if arr[node] == color {
return false
}
if arr[node] == 0 && !dfs(node, -color) {
return false
}
}
return true
}
for i := 0; i < len(arr); i++ {
if arr[i] == 0 {
if !dfs(i, 1) {
return false
}
}
}
return true
}