车的可用捕获量

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:

车的可用捕获量_第1张图片

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:

车的可用捕获量_第2张图片

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:

车的可用捕获量_第3张图片

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。
#获取到车的位置,直接在4个方向进行遍历
class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        if not board:
            return 0
        #找到车的起始位置
        for i,line in enumerate(board):
            for j,w in enumerate(line):
                if w == "R":
                    x,y = i,j
        count = 0
        #沿着一个方向运动,直到到达边界或者象、卒
        for (dx,dy) in [(-1,0),(1,0),(0,1),(0,-1)]:
            new_x = x+dx
            new_y = y+dy
            while not (new_x < 0 or new_x>=len(board) or new_y < 0 or new_y>=len(board[0])):
                if board[new_x][new_y] == "B":
                    break
                if board[new_x][new_y] == "p":
                    count += 1
                    break
                new_x += dx
                new_y += dy
        return count

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