前端算法第一七八期-可以被一步捕获的棋子数

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在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。

车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：

• 棋手选择主动停下来。
• 棋子因到达棋盘的边缘而停下。
• 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
• 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。

你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。

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。

模拟

根据题意模拟即可：

• 遍历棋盘确定白色车的下标，用 (st,ed)(st,ed)(st,ed) 表示。
• 模拟车移动的规则，朝四个基本方向移动，直到碰到卒或者白色象或者碰到棋盘边缘时停止，用 cnt\textit{cnt}cnt 记录捕获到的卒的数量。

那么如何模拟车移动的规则呢？我们可以建立方向数组表示在这个方向上移动一步的增量，比如向北移动一步的时候，白色车的 x 轴坐标减 1，而 y 轴坐标不会变化，所以我们可以用 (-1, 0) 表示白色车向北移动一步的增量，其它三个方向同理。建立了方向数组，则白色车在某个方向移动 $\textit{step}$ 步的坐标增量就可以直接计算得到，比如向北移动 $\textit{step}$ 步的坐标增量即为 (-step, 0)。

var numRookCaptures = function(board) {
    let cnt = 0, st = 0, ed = 0;
    const dx = [0, 1, 0, -1];
    const dy = [1, 0, -1, 0];

    for (let i = 0; i < 8; ++i) {
        for (let j = 0; j < 8; ++j) {
            if (board[i][j] == 'R') {
                st = i;
                ed = j;
                break;
            }
        }
    }
    for (let i = 0; i < 4; ++i) {
        for (let step = 0;; ++step) {
            const tx = st + step * dx[i];
            const ty = ed + step * dy[i];
            if (tx < 0 || tx >= 8 || ty < 0 || ty >= 8 || board[tx][ty] == 'B') {
                break;
            }
            if (board[tx][ty] == 'p') {
                cnt++;
                break;
            }
        }
    }
    return cnt;
};