785.Is-Graph-Bipartite?

785. Is Graph Bipartite?

题目地址

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

题目描述

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].

代码

Approach #1 Coloring by DFS

Complexity Analysis

• Time Complexity: O(N + E), where NN is the number of nodes in the graph, and E is the number of edges. We explore each node once when we transform it from uncolored to colored, traversing all its edges in the process.

• Space Complexity: O(N), the space used to store the color.

class Solution {
  public boolean isBipartite(int[][] graph) {
        int n = graph.length;
    int[] color = new int[n];
    Arrays.fill(color, -1);

    for (int start = 0; start < n; start++) {
      if (color[start] == -1) {
        Stack<Integer> stack = new Stack();
        stack.push(start);
        // color[start] = 0;    // whatever, It doesn't matter whether you add it or not

        while (!stack.isEmpty()) {
          Integer node = stack.pop();
          for (int nei: graph[node]) {
            if (color[nei] == -1) {
              stack.push(nei);
              color[nei] = color[node] ^ 1; // -1 ^ 1 = -2, -2 ^ 1 = -1
            } else if (color[nei] == color[node]) {
              return false;
            }
          }
        }
      }
    }

    return true;
  }
}

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<int> colors(graph.size());
        for (int i = 0; i < graph.size(); ++i) {
            if (colors[i] == 0 && !valid(graph, 1, i, colors)) {
                return false;
            }
        }
        return true;
    }
    bool valid(vector<vector<int>>& graph, int color, int cur, vector<int>& colors) {
        if (colors[cur] != 0) return colors[cur] == color;
        colors[cur] = color;
        for (int i : graph[cur]) {
            if (!valid(graph, -1 * color, i, colors)) {
                return false;
            }
        }
        return true;
    }
};

Approach #2 BFS

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<int> colors(graph.size());
        for (int i = 0; i < graph.size(); ++i) {
            if (colors[i] != 0) continue;
            colors[i] = 1;
            queue<int> q{{i}};
            while (!q.empty()) {
                int t = q.front(); q.pop();
                for (auto a : graph[t]) {
                    if (colors[a] == colors[t]) return false;
                    if (colors[a] == 0) {
                        colors[a] = -1 * colors[t];
                        q.push(a);
                    }
                }
            }
        }
        return true;
    }
};

Approch #3 Union Find

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<int> root(graph.size());
        for (int i = 0; i < graph.size(); ++i) root[i] = i;
        for (int i = 0; i < graph.size(); ++i) {
            if (graph[i].empty()) continue;
            int x = find(root, i), y = find(root, graph[i][0]);
            if (x == y) return false;
            for (int j = 1; j < graph[i].size(); ++j) {
                int parent = find(root, graph[i][j]);
                if (x == parent) return false;
                root[parent] = y;
            }
        }
        return true;
    }
    int find(vector<int>& root, int i) {
        return root[i] == i ? i : find(root, root[i]);
    }
};