LeetCode999-可以被一步捕获的棋子数(python)

该博客讨论了一个8x8棋盘上的国际象棋问题,涉及一个白色的车(Rook)如何按照规则捕获敌方卒(pawn)。在给定的棋盘配置示例中,算法需计算车能捕获的卒的数量,同时考虑不能进入有友方象(Bishop)的方格。博客提供了解决此问题的官方解决方案和一个实例解析,展示了如何模拟车的移动以计算捕获卒的数目。

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在一个 8 x 8 的棋盘上,有一个白色的车(Rook),用字符 ‘R’ 表示。棋盘上还可能存在空方块,白色的象(Bishop)以及黑色的卒(pawn),分别用字符 ‘.’,‘B’ 和 ‘p’ 表示。不难看出,大写字符表示的是白棋,小写字符表示的是黑棋。

车按国际象棋中的规则移动。东,西,南,北四个基本方向任选其一,然后一直向选定的方向移动,直到满足下列四个条件之一:

棋手选择主动停下来。
棋子因到达棋盘的边缘而停下。
棋子移动到某一方格来捕获位于该方格上敌方(黑色)的卒,停在该方格内。
车不能进入/越过已经放有其他友方棋子(白色的象)的方格,停在友方棋子前。
你现在可以控制车移动一次,请你统计有多少敌方的卒处于你的捕获范围内(即,可以被一步捕获的棋子数)。

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’

【官解】
根据题意模拟即可:

遍历棋盘确定白色车的下标,用 (st,ed)(st,ed) 表示。

模拟车移动的规则,朝四个基本方向移动,直到碰到卒或者白色象或者碰到棋盘边缘时停止,用cnt 记录捕获到的卒的数量。

那么如何模拟车移动的规则呢?我们可以建立方向数组表示在这个方向上移动一步的增量,比如向北移动一步的时候,白色车的 x 轴坐标减 1,而 y 轴坐标不会变化,所以我们可以用 (-1, 0) 表示白色车向北移动一步的增量,其它三个方向同理。建立了方向数组,则白色车在某个方向移动 step 步的坐标增量就可以直接计算得到,比如向北移动 step 步的坐标增量即为 (-step, 0)。

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        cnt, st, ed = 0, 0, 0
        dx, dy = [0, 1, 0, -1],[1, 0, -1, 0]
        for i in range(8):
            for j in range(8):
                if board[i][j] == "R":
                    st, ed = i, j
        for i in range(4):
            step = 0
            while True:
                tx = st + step * dx[i]
                ty = ed + step * dy[i]
                if tx < 0 or tx >= 8 or ty < 0 or ty >= 8 or board[tx][ty] == "B":
                    break
                if board[tx][ty] == "p":
                    cnt += 1
                    break
                step += 1
        return cnt

【个解】

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