public class Find2 {public int[] dx={1,-1,0,0};public int[] dy={0,0,1,-1};class Trie{Trie[] tries;String isEnd;public Trie(){tries = new Trie[26];}}public boolean[][] vis;    //是否判断过public List<String> res;    //答案public void insert(String word, Trie root){Trie t = root;for(int i = 0;i < word.length();i++){int index = word.charAt(i)-'a';if(t.tries[index] == null){t.tries[index] = new Trie();}t = t.tries[index];//跳到子节点}t.isEnd = word;}public List<String> findWords(char[][] board, String[] words) {//先把单词存入字典树当中Trie root=new Trie();for(String word:words){insert(word,root);}res=new ArrayList<String>();vis=new boolean[board.length][board[0].length];for(int i=0;i<board.length;++i){  //对board每一个点都进行检索for(int j=0;j<board[i].length;++j){dfs(root,i,j,board);}}Collections.sort(res);  //需要对结果进行排序return new ArrayList<String>(res);}public void dfs(Trie cur,int x,int y,char[][] board){//判断边界if(x<0||y<0||x>=board.length||y>=board[0].length||vis[x][y]){return;}cur=cur.tries[board[x][y]-'a'];  //延伸下一个节点vis[x][y]=true; //把当前设置为走过 不可重复走if(cur!=null){ //如果当前不为null的话 可以继续检索if(cur.tries!=null){ //说明到这里已经可以组成一个单词了res.add(cur.isEnd);cur.tries=null; //变成null是为了防止重复加入单词}for(int i=0;i<4;++i){dfs(cur,x+dx[i],y+dy[i],board); //四个方向检索}}vis[x][y]=false;}public static void main(String[] args) {String[] words ={"oath","pea","eat","rain"};char[][] board = {{'o', 'a', 'a', 'n'},{'e', 't', 'a', 'e'},{'i', 'h', 'k', 'r'},{'i', 'f', 'l', 'v'}};Solution m = new Solution();List<String> a = m.findWords(board,words);System.out.println(a.toString());}
}