README.md

题目描述: 车的可用捕获量

在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。

示例 :

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。

解题思路:

直接模拟法

代码

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int n = board.size();
        int res = 0;
        for (int i=0;i<n;i++)
        {
            for (int j=0;j<n;j++)
            {
                //找到R的位置
                if (board[i][j] == 'R')
                {
                    //以R 为原点建立坐标系
                    //依次向上找,向下找,向右找,向左找
                    res = cap(board, i, j, 0, 1) + cap(board, i, j, 0, -1) + cap(board, i, j, 1, 0) + cap(board, i, j, -1, 0);
                }
            }
        }
        return res;
    }

    int cap(vector<vector<char>> board, int x , int y, int dx, int dy)
    {
        /*参数说明 
         *a为原数组矩阵
         *x,y为R的坐标
         *dx,dy为增长步长
        */
        while(x>=0 && x<board.size() && y>=0 && y<board.size() && board[x][y]!='B'){
            if(board[x][y]=='p'){
                return 1;
            }
            x+=dx;
            y+=dy;
        }
        return 0;
    }
};