【LeetCode】999. 车的可用捕获量

题目

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 1：

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

示例 2：

输入：[[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。

示例 3：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 
车可以捕获位置 b5，d6 和 f5 的卒。

提示：

• board.length == board[i].length == 8
• board[i][j] 可以是 'R'，'.'，'B' 或 'p'
• 只有一个格子上存在 board[i][j] == 'R'

思路

每次可以在四个方向上分别移动一次，先找出'R'位置，然后在四个方向上移动。

代码

时间复杂度：O(n^2)
空间复杂度：O(1)

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        vector<int> dx = {0, 1, 0, -1}, dy = {1, 0, -1, 0};
        int tx, ty, res = 0;
        //1. 找出R位置
        for (int i = 0; i < 8; ++i) {
            for (int j = 0; j < 8; ++j) {
                if (board[i][j] == 'R') {
                    tx = i;
                    ty = j;
                    break;
                }
            }
        }
        //2. 分别朝四个方向上移动
        for (int i = 0; i < 4; ++i) {
            int step = 0;
            while (step < 8) {
                int x = tx + step * dx[i];
                int y = ty + step * dy[i];
                if (x < 0 || x >= 8 || y < 0 || y >= 8 || board[x][y] == 'B') break;
                if (board[x][y] == 'p') {
                    ++res;
                    break;
                }
                ++step;
            }
        }

        return res;

    }
};