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 contain u).

  • There are no parallel edges (graph[u] does not contain duplicate values).

  • If v is in graph[u], then u is in graph[v] (the graph is undirected).

  • The graph may not be connected, meaning there may be two nodes u and v such 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 == n

  • 1 <= n <= 100

  • 0 <= graph[u].length < n

  • 0 <= graph[u][i] <= n - 1

  • graph[u] does not contain u.

  • All the values of graph[u] are unique.

  • If graph[u] contains v, then graph[v] contains u.

解題

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
}

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