[LeetCode] 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.

DFS, 先枚举每个开始的位置，然后再用上下左右4个方向做DFS.

class Solution {
private:
    bool canUse[100][100];
    int step[4][2];
public:
    void check(int dep, int maxDep, string &word, vector<vector<char> > &board, bool &flag, int x, int y)
    {
        if (flag)
            return;
            
        if (dep == maxDep)
        {
            flag = true;
            return;
        }
        
        for(int i = 0; i < 4; i++)
        {
            int tx = x + step[i][0];
            int ty = y + step[i][1];
            if (0 <= tx && tx < board.size() && 0 <= ty && ty < board[0].size() && canUse[tx][ty] && 
                board[tx][ty] == word[dep])
            {
                canUse[tx][ty] = false;
                check(dep + 1, maxDep, word, board, flag, tx, ty);
                canUse[tx][ty] = true;
            }
        }
    }
    
    bool exist(vector<vector<char> > &board, string word) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (word.size() == 0)
            return true;
            
        memset(canUse, true, sizeof(canUse));
            
        step[0][0] = 1;
        step[0][1] = 0;
        step[1][0] = -1;
        step[1][1] = 0;
        step[2][0] = 0;
        step[2][1] = 1;
        step[3][0] = 0;
        step[3][1] = -1;
        
        for(int x = 0; x < board.size(); x++)
            for(int y = 0; y < board[x].size(); y++)
                if (board[x][y] == word[0])
                {
                    canUse[x][y] = false;
                    bool flag = false;
                    check(1, word.size(), word, board, flag, x, y);
                    if (flag)
                        return true;
                    canUse[x][y] = true;
                }
        
        return false;
    }
};