车的可用捕获量---JS

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/available-captures-for-rook
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每日一题，这道题挺简单的，但是一下子就能想到动行不动列，动列不动行，从中心往四个方向查找，但是当我在一个函数里去做操作的时候，运行超出时间限制。看了下题解和别人的写法，然后用JS写法来了，核心函数自己写没写好所以还是选择用题解给出的方案，走你。

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'

var numRookCaptures = function(board) {
     
   let arr = getR(board).split('')
   let Rrow = +arr[0]
   let Rcol = +arr[1]
   return getp(board,Rrow,Rcol,-1,0) + getp(board,Rrow,Rcol,1,0) + getp(board,Rrow,Rcol,0,-1) + getp(board,Rrow,Rcol,0,1)
};
function getp (arr,row,col,x,y) {
     
   while (row >= 0 && row < arr.length && col >= 0 && col < arr[row].length && arr[row][col] !== 'B') {
     
       if (arr[row][col] === 'p'){
     
           return 1
       }
       row+=x
       col+=y
   }
   return 0
}
function getR (board) {
     
   let Rrow = 0
   let Rcol = 0
   for(let i = 0; i < 8; i++) {
     
       for(let j = 0; j < 8; j++) {
     
           if ('R' === board[i][j]) {
     
               Rrow = i
               Rcol = j
           }
       }
   }
   return Rrow + '' + Rcol 
}