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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

没啥特别的思路，就是DFS全部暴搜一遍。

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        int rows = board.size();
        if (rows == 0) return false;
        int cols = board[0].size();
        if (cols == 0) return false;
        
        vector<vector<bool>> flag(rows, vector<bool>(cols, true));
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < cols; ++j) {
                if (dfs(board, flag, word, 0, i, j)) return true;
            }
        }
        
        return false;
    }
    
private:
    bool dfs(vector<vector<char>> &b, vector<vector<bool>> &flag, string &w, int cpos, int row, int col) {
        if (cpos == w.length()) return true;
        if (row < 0 || row >= b.size() || col < 0 || col >= b[0].size()) return false;
        if (!flag[row][col] || b[row][col] != w[cpos]) return false;
        flag[row][col] = false;
        if (dfs(b, flag, w, cpos + 1, row + 1, col) || dfs(b, flag, w, cpos + 1, row - 1, col) ||
            dfs(b, flag, w, cpos + 1, row, col + 1) || dfs(b, flag, w, cpos + 1, row, col - 1)) return true;
        flag[row][col] = true;
        return false;
    }
};