1. 题目描述

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：

image

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

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'

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

2. 解题思路

这道题思路简单，只需要找到车，然后向东南西北四个方向上遍历一下就行，如果遍历的过程中碰到空白就继续，碰到象就退出，碰到卒就计数器加一再推出。

2.1 代码

代码比较冗余，但是没什么优化的动力~

/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function(board) {
    // if (!board.length || !board[0].length) return 0
    let res = 0
    for (let i = 0; i < board.length; i++) {
        for (let j = 0; j < board[0].length; j++) {
            if (board[i][j] === 'R') {
                console.log(i, j)
                for (let m = i+1; m < board.length; m++) {
                    if (board[m][j] === '.') continue
                    if (board[m][j] === 'B') break
                    if (board[m][j] === 'p') {
                        res += 1
                        break
                    }
                }
                for (let m = i-1; m > -1; m--) {
                    if (board[m][j] === '.') continue
                    if (board[m][j] === 'B') break
                    if (board[m][j] === 'p') {
                        res += 1
                        break
                    }
                }
                for (let n = j+1; n < board[0].length; n++) {
                    if (board[i][n] === '.') continue
                    if (board[i][n] === 'B') break
                    if (board[i][n] === 'p') {
                        res += 1
                        break
                    }
                }
                for (let n = j-1; n > -1; n--) {
                    if (board[i][n] === '.') continue
                    if (board[i][n] === 'B') break
                    if (board[i][n] === 'p') {
                        res += 1
                        break
                    }
                }
                break
            }
        }
    }
    return res
};