LeetCode-999. Available Captures for Rook（车的可用捕获量）

车的可用捕获量

Available Captures for Rook

解

题目本身很简单，但是要注意题目描述的理解“返回车能够在一次移动中捕获到的卒的数量”
直接模拟就可以。

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int count = 0, x = 0, y = 0;
        int dx[4] = { 0, 1, 0, -1 };
        int dy[4] = { 1, 0, -1, 0 };
        for (int i = 0; i < 8; i++) {
            for (int j = 0; j < 8; j++) {
                if (board[i][j] == 'R') {
                    x = i;
                    y = j;
                    break;
                }
            }
        }
        for (int i = 0; i < 4; i++) {
            for (int j = 0;; j++) {
                int tx = x + j * dx[i];
                int ty = y + j * dy[i];
                if (tx < 0 || tx >= 8 || ty < 0 || ty >= 8 || board[tx][ty] == 'B') {
                    break;
                }
                if (board[tx][ty] == 'p') {
                    count++;
                    break;
                }
            }
        }
        return count;
    }
};