剑指 Offer 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

链接：https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof
思路：DFS + 剪枝

func exist(board [][]byte, word string) bool {
    row := len(board)
    col := len(board[0])
    for i := 0; i < row; i++ {
        for j := 0; j < col; j++ {
            if dfs(board, word, i, j, 0) {
                return true
            }
        }
    }
    return false
}

func dfs(board [][]byte, word string, row int, col int, t int) bool {
    if row < 0 || row > len(board) - 1 ||
        col < 0 || col > len(board[0]) - 1 ||
        board[row][col] != word[t] {
        return false
    }

    if t == len(word) - 1 {
        return true
    }

    t++
    var temp byte
    board[row][col], temp = '/', board[row][col]
    ret := dfs(board, word, row + 1, col, t) ||
           dfs(board, word, row - 1, col, t) ||
           dfs(board, word, row, col + 1, t) ||
           dfs(board, word, row, col - 1, t)

    board[row][col] = temp

    return ret
}