leetcode中如何解决车的可用捕获量问题
小编给大家分享一下leetcode中如何解决车的可用捕获量问题，希望大家阅读完这篇文章之后都有所收获，下面让我们一起去探讨吧！
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：
题目的意思是中间R走一下能吃p的次数，方向是上下左右四个方向
限制条件是走一次，以及遇到B就表明该方向不通
class Solution: def numRookCaptures(self, board: List[List[str]]) -> int: cnt, st, ed = 0, 0, 0 # 方向数组 direction = [(0, 1), (0, -1), (1, 0), (-1, 0)] # 找到R for i in range(8): for j in range(8): if board[i][j] == 'R': st = i ed = j # 朝着四个方向探索 for i in range(4): step = 0 while True: dx, dy = direction[i] tx = st + step * dx ty = ed + step * dy 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 #每次移动的范围，第一次移动一格，找不到的话移动两格 return cnt
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