999. Available Captures for Rook

问题：

给定国际象棋二维数组，

B代表阻止，p代表可吃点，R代表起始点。

从R开始向上下左右直线移动，遇到p点，res+1。遇到B或者棋盘边界res不增不减。

求最后res的值。

Example 1:
Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation: 
In this example the rook is able to capture all the pawns.

Example 2:
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.

Example 3:
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.

Note:
board.length == board[i].length == 8
board[i][j] is either 'R', '.', 'B', or 'p'
There is exactly one cell with board[i][j] == 'R'

　　

  

解法：

先遍历二维数组，找到R

然后向四个方向找到 p：res+1

找到 边界 or B 停止寻找。

代码参考：

class Solution {
public:
    int findp(vector<vector<char>>& board, int i, int j, int pi, int pj){
        while(i>=0 && i<board.size() && j>=0 && j<board[0].size()
              && board[i][j] != 'B'){
            if(board[i][j]=='p') return 1;
            i+=pi;
            j+=pj;
        }
        return 0;
    }
    
    int numRookCaptures(vector<vector<char>>& board) {
        int res=0;
        int n=board.size(), m=board[0].size();
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(board[i][j]=='R'){
                    res=findp(board, i, j, 1, 0)+findp(board, i, j, -1, 0)+
                        findp(board, i, j, 0, 1)+findp(board, i, j, 0, -1);
                }
            }
        }
        return res;
    }
};