999. Available Captures for Rook

[Tcdian]

999. Available Captures for Rook

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。

Example 1

Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation: 
In this example the rook is able to capture all the pawns.

Example 2

Input: [[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 0
Explanation: 
Bishops are blocking the rook to capture any pawn.

Example 3

Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation: 
The rook can capture the pawns at positions b5, d6 and f5.

Note

• board.length == board[i].length == 8
• board[i][j] is either 'R', '.', 'B', or 'p'
• There is exactly one cell with board[i][j] == 'R'

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function(board) {
    const direction = [-1, 0, 1, 0, -1];
    let result = 0;
    let Rx;
    let Ry;
    
    for (let i = 0; i < board.length; i++) {
        for (let j = 0; j < board[0].length; j++) {
            if (board[i][j] === 'R') {
                Rx = i;
                Ry = j;
                break;
            }        
        }
    }
    
    for (let d = 0; d < 4; d++) {
        let step = 1;
        while (true) {
            const dx = Rx + step * direction[d];
            const dy = Ry + step * direction[d + 1];
            if (dx < 0 || dx >= 8 || dy < 0 || dy >= 8 || board[dx][dy] === 'B') {
                break;
            }
            if (board[dx][dy] === 'p') {
                result++;
                break;
            }
            step++;
        }
    }
    
    return result;
    
};