【LeetCode】999. 車的可用捕獲量

題目鏈接：

999. 車的可用捕獲量

題目描述：

在一個 8 x 8 的棋盤上，有一個白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字符 “R”，“.”，“B” 和 “p” 給出。大寫字符表示白棋，小寫字符表示黑棋。

車按國際象棋中的規則移動：它選擇四個基本方向中的一個（北，東，西和南），然後朝那個方向移動，直到它選擇停止、到達棋盤的邊緣或移動到同一方格來捕獲該方格上顏色相反的卒。另外，車不能與其他友方（白色）象進入同一個方格。

返回車能夠在一次移動中捕獲到的卒的數量。

示例 1：

輸入：
[[".",".",".",".",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 [".",".",".","R",".",".",".","p"],
 [".",".",".",".",".",".",".","."],
 [".",".",".",".",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 [".",".",".",".",".",".",".","."],
 [".",".",".",".",".",".",".","."]]
輸出：3
解釋：在本例中，車能夠捕獲所有的卒。

示例 2：

輸入：
[[".",".",".",".",".",".",".","."],
 [".","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",".","."],
 [".",".",".",".",".",".",".","."],
 [".",".",".",".",".",".",".","."]]
輸出：0
解釋：象阻止了車捕獲任何卒。

示例 3：

輸入：
[[".",".",".",".",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 ["p","p",".","R",".","p","B","."],
 [".",".",".",".",".",".",".","."],
 [".",".",".","B",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 [".",".",".",".",".",".",".","."]]
輸出：3
解釋：車可以捕獲位置 b5，d6 和 f5 的卒。

提示：

1. board.length == board[i].length == 8
2. board[i][j] 可以是 'R'，'.'，'B' 或 'p'
3. 只有一個格子上存在 board[i][j] == 'R'

思路：

這道題很無聊，一句話形容這道題目：老太太的裹腳布——又臭又長。

題目本身非常簡單，但是題目描述讓人一言難盡。

題目的意思是，棋盤中有一個“車”，問“車”向上下左右四個方向遍歷能喫到多少個“卒”。條件是：

1. 不能出棋盤；
2. 遇到“象”不通；
3. 一旦喫到“卒”了，這個方向上的遍歷就結束了。

一旦描述清楚，題目就變得非常簡單了。也懶得想了，直接簡單粗暴吧。

class Solution {
    public int numRookCaptures(char[][] board) {
        // 結果數
        int count = 0;
        // 車的座標
        int rooki = 0, rookj = 0;
        // 找到車的位置
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                // 題目中已經說明只有一個“車”，所以一旦找到就可以終止尋找了。
                // 但棋盤只有8*8，跳不跳循環也沒啥區別。
                if (board[i][j] == 'R') {
                    rooki = i;
                    rookj = j;
                }
            }
        }

        // 向左
        for (int j = rookj; j >= 0; j--) {
            if (board[rooki][j] == 'B') {
                break;
            }
            if (board[rooki][j] == 'p') {
                count ++;
                break;
            }
        }

        // 向右
        for (int j = rookj; j < board[rooki].length; j++) {
            if (board[rooki][j] == 'B') {
                break;
            }
            if (board[rooki][j] == 'p') {
                count ++;
                break;
            }
        }

        // 向上
        for (int i = rooki; i >= 0; i--) {
            if (board[i][rookj] == 'B') {
                break;
            }
            if (board[i][rookj] == 'p') {
                count ++;
                break;
            }
        }

        // 向下
        for (int i = rooki; i < board.length; i++) {
            if (board[i][rookj] == 'B') {
                break;
            }
            if (board[i][rookj] == 'p') {
                count ++;
                break;
            }
        }

        return count;
    }
}