[Leetcode][第785题][JAVA][判断二分图][BFS][DFS]

àì夳堔傛蜴生んèń 2023-02-26 15:26 22阅读 0赞

#### 【问题描述】\[中等\] ####

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70]

#### 【解答思路】 ####

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 1]

###### 1. DFS 深度优先遍历 ######

**时间复杂度：O(N+M) 空间复杂度：O(N)**  
![在这里插入图片描述][20200716101907439.png]

class Solution { 
        private static final int UNCOLORED = 0;
        private static final int RED = 1;
        private static final int GREEN = 2;
        private int[] color;
        private boolean valid;
    
        public boolean isBipartite(int[][] graph) { 
            int n = graph.length;
            valid = true;
            color = new int[n];
            Arrays.fill(color, UNCOLORED);
            for (int i = 0; i < n && valid; ++i) { 
                if (color[i] == UNCOLORED) { 
                    dfs(i, RED, graph);
                }
            }
            return valid;
        }
    
        public void dfs(int node, int c, int[][] graph) { 
            color[node] = c;
            int cNei = c == RED ? GREEN : RED;
            for (int neighbor : graph[node]) { 
                if (color[neighbor] == UNCOLORED) { 
                    dfs(neighbor, cNei, graph);
                    if (!valid) { 
                        return;
                    }
                } else if (color[neighbor] != cNei) { 
                    valid = false;
                    return;
                }
            }
        }
    }

###### 2. 广度优先遍历 BFS ######

**时间复杂度：O(N+M) 空间复杂度：O(N)**  
![在这里插入图片描述][2020071610194691.png]

class Solution { 
        private static final int UNCOLORED = 0;
        private static final int RED = 1;
        private static final int GREEN = 2;
        private int[] color;
    
        public boolean isBipartite(int[][] graph) { 
            int n = graph.length;
            color = new int[n];
            Arrays.fill(color, UNCOLORED);
            for (int i = 0; i < n; ++i) { 
                if (color[i] == UNCOLORED) { 
                    Queue<Integer> queue = new LinkedList<Integer>();
                    queue.offer(i);
                    color[i] = RED;
                    while (!queue.isEmpty()) { 
                        int node = queue.poll();
                        int cNei = color[node] == RED ? GREEN : RED;
                        for (int neighbor : graph[node]) { 
                            if (color[neighbor] == UNCOLORED) { 
                                queue.offer(neighbor);
                                color[neighbor] = cNei;
                            } else if (color[neighbor] != cNei) { 
                                return false;
                            }
                        }
                    }
                }
            }
            return true;
        }
    }

#### 【总结】 ####

###### 1. 深度优先遍历 栈/递归 广度优先遍历 队列 ######

###### 2有向图、无向图 、树 ----> BFS DFS思想 ######

转载链接：https://leetcode-cn.com/problems/is-graph-bipartite/solution/pan-duan-er-fen-tu-by-leetcode-solution/

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70]: /images/20230209/a02992d4f7ed438cb2d6129be42df7d2.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 1]: https://img-blog.csdnimg.cn/20200716101759209.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk=,size_16,color_FFFFFF,t_70
[20200716101907439.png]: https://img-blog.csdnimg.cn/20200716101907439.png
[2020071610194691.png]: https://img-blog.csdnimg.cn/2020071610194691.png