79. 单词搜索

题目链接：leetcode.

从每一个位置开始，搜索是否能匹配

/*
执行用时：452 ms, 在所有 C++ 提交中击败了28.00%的用户
内存消耗：7.1 MB, 在所有 C++ 提交中击败了71.64%的用户
*/
class Solution {
	vector<vector<int>> visit;
	
	bool dfs(vector<vector<char>>& board, int i, int j, string word, int pos)
	{
        if(i < 0 || j < 0 || i >= board.size() || j >= board[0].size() || visit[i][j])
		{
			return false;
		} 
		if(board[i][j] != word[pos])
			return false;
		if(pos == word.size() - 1)
			return true;
		visit[i][j] = 1;
		bool b1 = dfs(board, i-1, j, word, pos+1), b2 = dfs(board, i+1, j, word, pos+1);
		bool b3 = dfs(board, i, j-1, word, pos+1), b4 = dfs(board, i, j+1, word, pos+1);
		visit[i][j] = 0;//递归回溯 
		return b1 || b2 || b3 || b4;
	} 
public:
    bool exist(vector<vector<char>>& board, string word) {
        int M = board.size(), N = board[0].size();
    	visit.resize(M, vector<int>(N));
    	for(int i = 0;i < M;++i)
    	{
    		for(int j = 0;j < N;++j)
    		{
    			if(dfs(board,i,j,word,0))
    			{
    				return true;
    			}
    		}
    	}
    	return false;
    }
};

--------2021-05-12-------------------
重做后发现了一个神奇的事，就是方向矩阵如果用二维的话就会超时，改成一维的pair就没有问题，emmmm

/*
执行用时：324 ms, 在所有 C++ 提交中击败了53.55%的用户
内存消耗：7.3 MB, 在所有 C++ 提交中击败了30.22%的用户
*/

class Solution {
    // vector<vector<int>> dic = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    vector<pair<int, int>> dic{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    bool dfs(vector<vector<char>>& board, vector<vector<int>>& visit, string& word, int i, int j, int index)
    {
        if(board[i][j] != word[index])
        {
            return false;
        }
        else if(index == word.size() - 1)
            return true;
        char tmp = board[i][j];
        visit[i][j] = 1;//标记已访问
        int M = board.size(), N = board[0].size();
        
        bool ans = false;
        for(auto x : dic)
        {
            int ti = i + x.first, tj = j + x.second;
            if(ti >= 0 && ti < M && tj >= 0 && tj < N && !visit[ti][tj])
            {
                if(dfs(board, visit, word, ti, tj, index + 1))
                {
                    ans = true;
                    break;
                }
            }
        }
        visit[i][j] = 0;
        return ans;
    }
public:
    bool exist(vector<vector<char>>& board, string word) {
        int M = board.size(), N = board[0].size();
        if(word.empty())
            return false;
        vector<vector<int>> visit(M, vector<int>(N, 0));
        for(int i = 0;i < M;++i)
        {
            for(int j = 0;j < N;++j)
            {
                if(board[i][j] == word[0] && dfs(board, visit, word, i, j, 0))
                    return true;
            }
        }
        return false;
    }
};