LeetCode 79. Word Search

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,

word = "SEE", -> returns true,

word = "ABCB", -> returns false.

枚举起点， dfs四个方向， 用一个标记数组标记一下走过的路径； 如果当前点可走，仅当当前点合法且未走过且等于string的下一个值。

代码如下：

class Solution {

    bool dfs(vector<vector<int>> &dir, vector<vector<char>> &board, vector<vector<int>> &mk, string &str, int x, int y, int n)
    {
        if(str.size() == n)
            return true;

        for(int i=0; i<dir.size(); ++ i)
        {
            int x1 = x + dir[i][0], y1 = y + dir[i][1];
            if(x1 >= 0 && y1 >= 0 && x1 < board.size() && y1 < board[0].size() && board[x1][y1] == str[n] && mk[x1][y1] == 0)
            {
                mk[x1][y1] = 1;
                if(dfs(dir, board, mk, str, x1, y1, n+1))
                    return true;
                mk[x1][y1] = 0;
            }
        }
        return false;
    }
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(board.size() == 0)
            return false;
        if(word.size() == 0)
            return true;

        vector<vector<int>> mk(board.size(), vector<int>(board[0].size(), 0));
        vector<vector<int>> dir{ {1, 0}, {-1, 0}, {0, 1}, {0, -1} };
        for(int i=0; i<board.size(); ++ i)
        {
            for(int j=0; j<board[0].size(); ++ j)
            {
                if(board[i][j] == word[0])
                {
                    mk[i][j] = 1;
                    if(dfs(dir, board, mk, word, i, j, 1))
                        return true;
                    mk[i][j] = 0;
                }
            }
        }
        return false;
    }
};