999. 车的可用捕获量 Golang 每日一题 扫描

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。

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

这两天的每日一题有点水。这道题限定了只有一个车，所有可以用这样的简单扫描。那如果是多个车呢？为了防止重复计数，我最开始想到的是用标记矩阵，后来又觉得用set更好，你觉得呢？

我的解答

func numRookCaptures(board [][]byte) int {
	for i:=0;i<8;i++{
		for j:=0;j<8;j++{
			if board[i][j]=='R'{
				return capture(board,i,j)
			}
		}
	}
    return 0
}
func capture(board [][]byte,i,j int) int {
	d:=[][]int{{0,1},{0,-1},{1,0},{-1,0}}
	cnt:=0
	for k:=0;k<4;k++{
		x,y:=i,j
		for x+d[k][0]>=0 &&  y+d[k][1] >=0 && x+d[k][0]<8 &&  y+d[k][1] <8 {
			x,y=x+d[k][0] ,  y+d[k][1]
			if board[x][y]=='p'{
				cnt++
				break
			}else if board[x][y]=='B'{
				break
			}
		}
	}
	return cnt
}