题目：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。
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
- 模拟题，按照题目描述做就好
- 首先定义一个方向数组，遍历四个方向累计得到结果。
实现
func numRookCaptures(board [][]byte) int {
var ans int
dt := [4][2]int{{-1, 0}, {1, 0}, {0, 1}, {0, -1}}
for row := 0; row < 8; row++ {
for col := 0; col < 8; col++ {
if board[row][col] == 'R' {
for _, d := range dt {
ans += _impel(board, row, col, d)
}
break
}
}
}
return ans
}
func _impel(board [][]byte, x, y int, d [2]int) int {
for {
x += d[0]
y += d[1]
if x < 0 || x >= 8 || y < 0 || y >= 8 {
break
}
if board[x][y] == 'B' {
return 0
} else if board[x][y] == 'p' {
return 1
}
}
return 0
}
