解法 典型的深度优先遍历,注意isVisted的置1和置0 代码 class Solution { public boolean exist(char[][] board, String word) { int[][] isVisited = new int[board.length][board[0].length]; for (int i = 0; i < board.length; i++) { for (int j = 0; j < board[0].length; j++) { if(f(board,isVisited,word,0,i,j)) return true; ...