999. 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 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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/available-captures-for-rook
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题解

以R为中心点十字搜索

/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function(board) {
    let r = [], res = 0, s = ''
    for (let i = 0; i < board.length; i++) {
        if (board[i].indexOf('R') != -1) {
            r = [i, board[i].indexOf('R')]
            break
        }
    }
    for (let i = 0; i < board.length; i++) {
        s += board[i][r[1]]
    }
    if (/p[^B]*R/.test(board[r[0]].join(''))) {
        res++
    }
    if (/R[^B]*p/.test(board[r[0]].join(''))) {
        res++
    }
    if (/p[^B]*R/.test(s)) {
        res++
    }
    if (/R[^B]*p/.test(s)) {
        res++
    }
    return res
};