原题链接：

886. 可能的二分法
https://leetcode.cn/problems/possible-bipartition/description/

完成情况：

题解一：

package LeetCode中等题;import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;public class __886可能的二分法__广度优先搜索 {/*** @param n* @param dislikes* @return*/public boolean possibleBipartition(int n, int[][] dislikes) {int color [] = new int[n+1];List<Integer> [] g_dyeing = new List[n+1];for (int i=0;i<=n;i++){g_dyeing[i] = new ArrayList<Integer>();}for (int p[] :dislikes){g_dyeing[p[0]].add(p[1]);g_dyeing[p[1]].add(p[0]);}for (int i=1;i<=n;i++){if (color[i] == 0){Queue<Integer> queue = new ArrayDeque<Integer>();queue.offer(i);color[i] = 1;while (!queue.isEmpty()){int t = queue.poll();for (int next:g_dyeing[t]){if (color[next] > 0 && color[next] == color[t]){return false;}if (color[next] == 0){color[next] = 3^color[t];queue.offer(next);}}}}}return true;}
}

题解二：

package LeetCode中等题;import java.util.ArrayList;
import java.util.List;public class __886可能的二分法__深度优先搜索__图的染色法 {/**输入：n = 4, dislikes = [[1,2],[1,3],[2,4]]n代表n个人，dislike中数组表示指定两个不能配在一起。问：：能否找到满足条件的情况？* @param n* @param dislikes* @return*/public boolean possibleBipartition(int n, int[][] dislikes) {int color [] = new int[n+1];List<Integer> [] g_dyeing = new List[n+1];for (int i=0;i<=n;i++){g_dyeing[i] = new ArrayList<Integer>();}for (int p[] :dislikes){g_dyeing[p[0]].add(p[1]);g_dyeing[p[1]].add(p[0]);}for (int i=1;i<=n;i++){if (color[i] == 0 && !dfs_possibleBipartition(i,1,color,g_dyeing)){return false;}}return true;}/**** @param curNode* @param nowColor* @param color* @param g_dyeing* @return*/private boolean dfs_possibleBipartition(int curNode, int nowColor, int[] color, List<Integer>[] g_dyeing) {color[curNode] = nowColor;for (int nextNode : g_dyeing[curNode]){if (color[nextNode] != 0 && color[nextNode] == color[curNode]){return false;}if (color[nextNode] == 0 && !dfs_possibleBipartition(nextNode,3^nowColor,color,g_dyeing)){return false;}}return true;}
}