Leetcode 419 Battleships in a Board

解题思路:

题目大意是战舰只能一行或者一列这样的排列,并且战舰必须隔开,不能有相交.

所以只要遍历是'X'的点,并且'X'的上方和左侧不能是'X',这样就是一个战舰.

class Solution {public:    bool is_X(int i,int j,vector<vector<char> >& board){        if (board[i][j] != 'X')            return false;        else{            if (i==0 && j==0)                return true;            if (i==0 && j>0 && board[i][j-1] != 'X')                return true;            if (i>0 && j==0 && board[i-1][j] != 'X')                return true;            if (i > 0 && j > 0 && board[i-1][j] != 'X' && board[i][j-1] != 'X')                return true;            else                return false;    }    }    int countBattleships(vector<vector<char> >& board) {        int res = 0;        int m = board.size();        int n = board[0].size();        for (int i = 0; i < m; i++)            for(int j = 0; j < n; j++){                if (is_X(i,j,board) == true)                    res++;            }        return res;    }};

也可以使用dfs,

class Solution {public:    int m, n;     vector<vector<bool>> flag;    int move[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};    void dfs(vector<vector<char>>& board, int i, int j) {        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] == '.' || flag[i][j]) return;        flag[i][j] = true;        for (int d = 0; d < 4; ++d) dfs(board, i+move[d][0], j+move[d][1]);    }    int countBattleships(vector<vector<char>>& board) {        if (board.empty()) return 0;        m = board.size(), n = board[0].size();        flag.resize(m, vector<bool>(n, false));        int result = 0;        for (int i = 0; i < m; ++i)            for (int j = 0; j < n; ++j)                if (board[i][j] == 'X' && !flag[i][j]) {                    ++result;                    dfs(board, i, j);                }        return result;    }};