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785. Is Graph Bipartite?

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[LeetCode 시즌 3] 2022년 4월 28일 문제입니다.

https://leetcode.com/problems/is-graph-bipartite/


[Medium] 785. Is Graph Bipartite?

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.

관련자료

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austin님의 댓글

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class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<optional<bool>> party(graph.size());
        queue<pair<int, bool>> q;
        for(int i = 0; i < graph.size(); ++i) {
            if (party[i].has_value()) continue;
            q.emplace(i, true);
            while(!q.empty()) {
                auto[node, p] = q.front();
                q.pop();
                party[node] = p;
                for(auto c : graph[node]) {
                    if (!party[c].has_value()) q.emplace(c, !p);
                    else if (*party[c] == p) return false;
                }
            }
        }
        return true;
    }
};
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