＃886 Possible Bipartition

Description

We want to split a group of n people (labeled from 1 to n) into two groups of any size. Each person may dislike some other people, and they should not go into the same group.

Given the integer n and the array dislikes where dislikes[i] = [ai, bi] indicates that the person labeled ai does not like the person labeled bi, return true if it is possible to split everyone into two groups in this way.

Examples

Example 1:

Input: n = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4] and group2 [2,3].

Example 2:

Input: n = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:

Input: n = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

Constraints:

1 <= n <= 2000
0 <= dislikes.length <= 104
dislikes[i].length == 2
1 <= dislikes[i][j] <= n
ai < bi
All the pairs of dislikes are unique.

思路

当然他和785很接近，所以最开始的时候，就用BFS去判断能不能2色填充。

但是我在submission里面看到了另一种用并查集的方式，其实我不知道它的原理是什么，只是感官上觉得这种方式是合理的，就先记录一下，以后可以直接用

代码

BFS

class Solution {
    
    public boolean possibleBipartition(int n, int[][] dislikes) {
    
        List<Set<Integer>> neighbors = new ArrayList<>();
        if (dislikes.length == 0)
            return true;
        
        for (int i = 0; i < n; i++)
            neighbors.add(new HashSet<>());
        
        for (int[] pairs: dislikes) {
    
            neighbors.get(pairs[0] - 1).add(pairs[1] - 1);
            neighbors.get(pairs[1] - 1).add(pairs[0] - 1);
        }
        
        int[] color = new int[n];
        int[] visited = new int[n];
        
        for (int i = 0; i < n ; i++) {
    
            if (visited[i] == 1)
                continue;
            
            List<Integer> parent = new ArrayList<>();
            parent.add(i);
            int currColor = 1;
            color[i] = 1;
            
            while(parent.size() != 0) {
    
                List<Integer> children = new ArrayList<>();
                for (int start: parent) {
    
                    if (visited[start] == 1)
                        continue;
                    visited[start] = 1;
                    for (int end: neighbors.get(start)) {
    
                        if (color[end] == currColor)
                            return false;
                        if (color[end] == -currColor)
                            continue;
                        color[end] = -currColor;
                        children.add(end);
                    }
                }
                currColor = -currColor;
                parent = children;
            }
        }
        
        
        return true;
    }
}

并查集

class Solution {
    
    int[] parent;
    public void union(int x, int y) {
    
        int parentX = find(x);
        int parentY = find(y);
        
        if (parentX == parentY)
            return;
        
        parent[parentX] = parentY;
    }
    
    public int find(int x) {
    
        if (x != parent[x])
            parent[x] = find(parent[x]);
        
        return parent[x];
    }
    
    public boolean possibleBipartition(int n, int[][] dislikes) {
    
        parent = new int[2 * n];
        for (int i = 0; i < n * 2; i++) {
    
            parent[i] = i;
        }
        
        for (int[] dislike: dislikes) {
    
            union(dislike[0] - 1 + n, dislike[1] - 1);
            union(dislike[0] - 1, dislike[1] - 1 + n);
        }
        
        for (int i = 0; i < n; i++) {
    
            if (find(i) == find(i + n))
                return false;
        }
        
        return true;
    }
}