车的可用捕获量
-题目-
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。
-方法1-
找到车之后，判断车所在行和列有没有符合条件的卒即可。有4个方向需要遍历。除了for循环，还可用一个数组标示遍历方向。
-ac代码-
class Solution:
def __init__(self):
self.board = None
def count(self, arr):
for i_, j_ in arr:
if self.board[i_][j_] == "B":
return 0
if self.board[i_][j_] == "p":
return 1
return 0
def count_available_captures(self, i, j):
return (
self.count([(i_, j) for i_ in range(i, 0, -1)])
+ self.count([(i_, j) for i_ in range(i, 8, 1)])
+ self.count([(i, j_) for j_ in range(j, 0, -1)])
+ self.count([(i, j_) for j_ in range(j, 8, 1)])
)
def numRookCaptures(self, board: List[List[str]]) -> int:
self.board = board
for i in range(len(board)):
for j in range(len(board)):
if board[i][j] == 'R':
return self.count_available_captures(i, j)
-复杂度-
- \(T(n) = O(n^2)\) （找车遍历\(n^2\)，找到后遍历\(n\)）
- \(S(n) = O(n)\) （存储方向数组）