面试题12. 矩阵中的路径

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

示例 2：

输入：board = [["a","b"],["c","d"]], word = "abcd"
输出：false

提示：
1 <= board.length <= 200
1 <= board[i].length <= 200

// 深度优先搜索 + 回溯
const exist = (board, word) => {

  const dfs = (board, word, i, j, k) => {
    // 处理边界
    if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] !== word[k] ) {
      return false;
    }
    // 深度结束标志
    if (k === word.length - 1) {
      return true;
    }

    const temp = board[i][j];
    board[i][j] = '/';
    const result = dfs(board, word, i-1, j, k+1) || dfs(board, word, i, j+1, k+1) || dfs(board, word, i+1, j, k+1) || dfs(board, word, i, j-1, k+1);
    board[i][j] = temp;

    return result;
  }

  for(let i = 0; i < board.length; i+=1) {
    for(let j = 0; j < board[0].length; j+=1) {
      // 一定要先匹配到了才开始dfs
      if (board[i][j] === word[0]) {
        if (dfs(board, word, i, j, 0)) {
          return true;
        }
      }
    }
  }

  return false;
}