999. Available Captures for Rook(Leetcode每日一题-2020.03.26)

Problem

On an 8 x 8 chessboard, there is one white rook. There also may be empty squares, white bishops, and black pawns. These are given as characters ‘R’, ‘.’, ‘B’, and ‘p’ respectively. Uppercase characters represent white pieces, and lowercase characters represent black pieces.

The rook moves as in the rules of Chess: it chooses one of four cardinal directions (north, east, west, and south), then moves in that direction until it chooses to stop, reaches the edge of the board, or captures an opposite colored pawn by moving to the same square it occupies. Also, rooks cannot move into the same square as other friendly bishops.

Return the number of pawns the rook can capture in one move.
Note:

1. board.length == board[i].length == 8
2. board[i][j] is either ‘R’, ‘.’, ‘B’, or ‘p’
3. There is exactly one cell with board[i][j] == ‘R’

Example1

Input: [[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",“R”,".",".",".",“p”],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation:
In this example the rook is able to capture all the pawns.

Example2

Input: [[".",".",".",".",".",".",".","."],[".",“p”,“p”,“p”,“p”,“p”,".","."],[".",“p”,“p”,“B”,“p”,“p”,".","."],[".",“p”,“B”,“R”,“B”,“p”,".","."],[".",“p”,“p”,“B”,“p”,“p”,".","."],[".",“p”,“p”,“p”,“p”,“p”,".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 0
Explanation:
Bishops are blocking the rook to capture any pawn.

Example3

Input: [[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",“p”,".",".",".","."],[“p”,“p”,".",“R”,".",“p”,“B”,"."],[".",".",".",".",".",".",".","."],[".",".",".",“B”,".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation:
The rook can capture the pawns at positions b5, d6 and f5.

Solution

这题的描述也是醉了，没看明白。
从题解，看到一个说法，可以理解为“车移动一次吃到卒的方案有多少种”。

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int ret = 0;
        if(board.empty())
            return ret;
        int row = board.size();
        int col = board[0].size();

        vector<pair<int,int>> dir={{-1,0},{1,0},{0,-1},{0,1}};

        for(int i = 0;i<row;++i)
        {
            for(int j = 0;j<col;++j)
            {
                if(board[i][j] == 'R')
                {
                    for(int k = 0;k<dir.size();++k)
                    {
                        int cur_r = i;
                        int cur_c = j;
                        while(1)
                        {
                            cur_r += dir[k].first;
                            cur_c += dir[k].second;
                            if(cur_r >= 0 && cur_r < row && cur_c >= 0 && cur_c < col)
                            {
                                if(board[cur_r][cur_c] == 'B' )
                                    break;
                                else if(board[cur_r][cur_c] == 'p')
                                {
                                    ++ret;
                                    break;
                                }
                            }
                            else
                            {
                                break;
                            }
                        }
                    }

                }
            }
        }
        return ret;

    }
};