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: trueExplanation: The graph looks like this:0----1|    ||    |3----2We 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: falseExplanation: The graph looks like this:0----1| \  ||  \ |3----2We cannot find a way to divide the set of nodes into two independent subsets.

 

M1: BFS

time: O(V + E), space: O(V)

class Solution {    public boolean isBipartite(int[][] graph) {        int[] visited = new int[graph.length];        for(int i = 0; i < graph.length; i++) {            if(!BFS(graph, i, visited)) {                return false;            }        }        return true;    }        private boolean BFS(int[][] graph, int node, int[] visited) {        if(visited[node] != 0) {            return true;        }        Queue
     
       q = new LinkedList<>();        q.offer(node);        visited[node] = 1;        while(!q.isEmpty()) {            int cur = q.poll();            int curColor = visited[cur];            int neiColor = -curColor;            for(int nei : graph[cur]) {                if(visited[nei] == 0) {                    visited[nei] = neiColor;                    q.offer(nei);                } else if(visited[nei] != neiColor) {                    return false;                }            }        }        return true;    }}
     

 

M2: DFS

Graph Coloring (by DFS/BFS): color a node as red, and then color all its neighbors as blue recursively. if there's any conflict, return false.

染色法 Graph Coloring (by DFS)

将相连的两个顶点染成不同的颜色，一旦在染的过程中发现有两连的两个顶点已经被染成相同的颜色，说明不是二分图。

-> 使用两种颜色，分别用1和-1来表示，初始时每个顶点用0表示未染色 

-> 遍历每一个顶点，如果该顶点未被访问过，调用递归函数，如果返回false，说明不是二分图，直接返回false；

     　　　　　　　　如果循环退出后没有返回false，则返回true。

-> 在递归函数中，如果当前顶点已经染色：如果该顶点的颜色和将要染的颜色相同，则返回true；否则返回false。

                          如果没被染色，则将当前顶点染色(-1*color: 将相邻点染成不同颜色)，然后再遍历与该顶点相连的所有的顶点，调用递归函数，如果返回false了，则当前递归函数的返回false，循环结束返回true.

 

time: O(V+E), space: O(V+E)

class Solution {    int[] colors;    public boolean isBipartite(int[][] graph) {        int n = graph.length;        colors = new int[n];        for(int i = 0; i < n; i++) {            if(colors[i] == 0 && !dfs(graph, i, 1))                return false;        }        return true;    }        private boolean dfs(int[][] graph, int node, int color) {        if(colors[node] != 0)            return colors[node] == color ? true: false;        else {            colors[node] = color;            for(int j : graph[node]) {                if(!dfs(graph, j, -color))                    return false;            }        }        return true;    }}