Leetcode daily 20/03/26 车的可用捕获量

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）