【Leetcode】79. Word Search

思路：

   遍历矩阵，结合dfs解即可。

   

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:

    Solution() {}

    bool exist(vector<vector<char>>& board, string word)
    {
        //初始化都没有被访问过
        this->rowCount = board.size();
        if (rowCount == 0) {
            return false;
        }
        this->colCount = board[0].size();
        for (int i = 0; i < rowCount; i++) {
            vector<int> v(colCount, 0);
            visit.push_back(v);
        }


        this->word = word;
        this->board = board;

        for (int i = 0; i < rowCount; i++){
            for (int j = 0; j< colCount; j++) {
                if (board[i][j] == word[0] && dfs(i, j, 0)) {
                    return true;
                }
            }
        }

        return false;
    }

    /**
     *@param row 当前字符所在行
     *@param col 当前字符所在列
     *@param index word当前被扫描的位置
     */
    bool dfs(int row, int col, int index)
    {
        if (index == word.size() - 1) {
            return true;
        }
        //当前位置标志为被访问
        visit[row][col] = 1;
        //up
        if (row - 1 >= 0 && !visit[row - 1][col] && board[row - 1][col] == word[index + 1] ) {
            if (dfs(row - 1, col, index + 1)) {
                return true;
            }
        }
        //down
        if (row + 1 < this->rowCount && !visit[row + 1][col] && board[row + 1][col] == word[index + 1]) {
            if (dfs(row + 1, col, index + 1)) {
                return true;
            }
        }
        //left
        if (col - 1 >= 0 && !visit[row][col - 1] && board[row][col - 1] == word[index + 1]) {
            if (dfs(row, col - 1, index + 1)) {
                return true;
            }
        }
        //right
        if (col + 1 < this->colCount && !visit[row][col + 1] && board[row][col + 1] == word[index + 1]) {
            if (dfs(row, col + 1, index + 1)) {
                return true;
            }
        }
        //不满足，置为0
        visit[row][col] = 0;
        return false;
    }

private:
    vector<vector<char>> board;
    string word;
    vector<vector<int>> visit;
    int rowCount;
    int colCount;
};

int main(int argc, char *argv[])
{
    vector<vector<char>> board =  {
            {'a','b'}};
    Solution s;
    string word = "ba";
    bool ret = s.exist(board, word);
    cout<<ret<<endl;
    return 0;
}