785. Is Graph Bipartite.md

785. Is Graph Bipartite?

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leetcode-cn Daily Challenge on July 16th, 2020.

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Difficulty : Medium

Related Topics : BFS、DFS、Graph

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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].

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Solution

• mine
  • Java
    • DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.9 MB, less than 78.94% of Java online submissions

      // O(N)time
      // O(N)space
      int[] colors;
      boolean res;
      public boolean isBipartite(int[][] graph) {
          res = true;
          int n = graph.length;
          colors = new int[n];
          for(int i = 0; i < n; i++){
              if(colors[i] != 0){
                  continue;
              }
              dfs(i, 1, graph);
          }
          return res;
      }
      
      void dfs(int index, int color, int[][] graph){
          colors[index] = color;
          color = color == 1 ? 2 : 1;
          for(int i : graph[index]){
              if(colors[i] == 0){
                  if(!res){
                      return;
                  }
                  dfs(i, color, graph);
              }else if(colors[i] != color){
                  res = false;
                  return;
              }
          }
      }
      

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• the most votes

  • DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.9 MB, less than 74.75% of Java online submissions

    // O(N)time
    // O(N)space
    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        int[] colors = new int[n];			
    			
        for (int i = 0; i < n; i++) {              //This graph might be a disconnected graph. So check each unvisited node.
            if (colors[i] == 0 && !validColor(graph, colors, 1, i)) {
                return false;
            }
        }
        return true;
    }
    
    public boolean validColor(int[][] graph, int[] colors, int color, int node) {
        if (colors[node] != 0) {
            return colors[node] == color;
        }       
        colors[node] = color;       
        for (int next : graph[node]) {
            if (!validColor(graph, colors, -color, next)) {
                return false;
            }
        }
        return true;
    }
    

  • BFS Runtime: 3 ms, faster than 30.19%, Memory Usage: 51.9 MB, less than 5.04% of Java online submissions

    // O(N)time
    // O(N)space
    public boolean isBipartite(int[][] graph) {
        int len = graph.length;
        int[] colors = new int[len];
        
        for (int i = 0; i < len; i++) {
            if (colors[i] != 0) continue;
            Queue<Integer> queue = new LinkedList<>();
            queue.offer(i);
            colors[i] = 1;   // Blue: 1; Red: -1.
            
            while (!queue.isEmpty()) {
                int cur = queue.poll();
                for (int next : graph[cur]) {
                    if (colors[next] == 0) {          // If this node hasn't been colored;
                        colors[next] = -colors[cur];  // Color it with a different color;
                        queue.offer(next);
                    } else if (colors[next] != -colors[cur]) {   // If it is colored and its color is different, return false;
                        return false;
                    }
                }
            }
        }
        
        return true;
    }
    

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