LeetCode 999. 车的可用捕获量（C++、python）

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：

LeetCode 999. 车的可用捕获量（C++、python）

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：

LeetCode 999. 车的可用捕获量（C++、python）

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'

C++

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) 
    {
        int res=0;
        int n=8;
        int row,col;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if('R'==board[i][j])
                {
                    row=i;
                    col=j;
                    break;
                }
            }
        }
        for(int i=col+1;i<n;i++)
        {
            if('B'==board[row][i])
            {
                break;
            }
            if('p'==board[row][i])
            {
                res++;
                break;
            }
        }
        for(int i=col-1;i>=0;i--)
        {
            if('B'==board[row][i])
            {
                break;
            }
            if('p'==board[row][i])
            {
                res++;
                break;
            }
        }
        for(int i=row-1;i>=0;i--)
        {
            if('B'==board[i][col])
            {
                break;
            }
            if('p'==board[i][col])
            {
                res++;
                break;
            }
        }
        for(int i=row+1;i<n;i++)
        {
            if('B'==board[i][col])
            {
                break;
            }
            if('p'==board[i][col])
            {
                res++;
                break;
            }
        }
        return res;
    }
};

python

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        n=8
        row=0
        col=0
        res=0
        for i in range(n):
            for j in range(n):
                if 'R'==board[i][j]:
                    row=i
                    col=j
                    break
        for i in range(col+1,n):
            if 'B'==board[row][i]:
                break
            if 'p'==board[row][i]:
                res+=1
                break
        for i in range(col-1,-1,-1):
            if 'B'==board[row][i]:
                break
            if 'p'==board[row][i]:
                res+=1
                break
        for i in range(row+1,n):
            if 'B'==board[i][col]:
                break
            if 'p'==board[i][col]:
                res+=1
                break
        for i in range(row-1,-1,-1):
            if 'B'==board[i][col]:
                break
            if 'p'==board[i][col]:
                res+=1
                break
        return res

LeetCode 999. 车的可用捕获量（C++、python）

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