Possible Bipartition

1、题目

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group. 

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

 

Example 1:

Input: N = 4, dislikes = [[1,2],[1,3],[2,4]] Output: true Explanation: group1 [1,4], 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

 

Note:

1 <= N <= 20000 <= dislikes.length <= 100001 <= dislikes[i][j] <= Ndislikes[i][0] < dislikes[i][1]There does not exist i != j for which dislikes[i] == dislikes[j].

   题目大意：从1到N将这堆数分成两个group,能否满足互相dislike的不分到同一个group里面。

2、思路

    二分图

3、代码

public boolean possibleBipartition(int N, int[][] dislikes) { List<Integer>[] graph = new List[N]; for (int i = 0; i < graph.length; i++) { graph[i] = new ArrayList<>(); } for (int i = 0; i < dislikes.length; i++) { graph[ dislikes[i][0] - 1 ].add( dislikes[i][1] - 1 ); graph[ dislikes[i][1] - 1 ].add( dislikes[i][0] - 1 ); } int[] colors = new int[N]; Arrays.fill(colors, -1); for (int i = 0; i < N; i++){ if (colors[i] == -1 && !paint(colors, i, graph, 0)){ return false; } } return true; } private boolean paint(int[] colors, int index, List<Integer>[] graph, int color) { colors[index] = color; for (int i = 0; i < graph[index].size(); i++){ int nextIndex = graph[index].get(i); if (colors[nextIndex] == color) { return false; } else if (colors[nextIndex] == -1 && !paint(colors, nextIndex, graph, 1 - color)) { return false; } } return true; }