public boolean exist(char[][] board, String word) {
char[] words = word.toCharArray();
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
//从[i,j]这个坐标开始查找
if (dfs(board, words, i, j, 0))
return true;
}
}
return false;
}
boolean dfs(char[][] board, char[] word, int i, int j, int index) {
//边界的判断，如果越界直接返回false。index表示的是查找到字符串word的第几个字符，
//如果这个字符不等于board[i][j]，说明验证这个坐标路径是走不通的，直接返回false
if (i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[index])
return false;
//如果word的每个字符都查找完了，直接返回true
if (index == word.length – 1)
return true;
//把当前坐标的值保存下来，为了在最后复原
char tmp = board[i][j];
//然后修改当前坐标的值
board[i][j] = ‘.’;
//走递归，沿着当前坐标的上下左右4个方向查找
boolean res = dfs(board, word, i + 1, j, index + 1) || dfs(board, word, i – 1, j, index + 1) ||
dfs(board, word, i, j + 1, index + 1) || dfs(board, word, i, j – 1, index + 1);
//递归之后再把当前的坐标复原
board[i][j] = tmp;
return res;
}