LeetCode 999 車的可用捕獲量

999. 車的可用捕獲量

在乙個 8 x 8 的棋盤上，有乙個白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字元「r」，「.」，「b」和「p」給出。大寫字元表示白棋，小寫字元表示黑棋。

車按西洋棋中的規則移動：它選擇四個基本方向中的乙個（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）象進入同乙個方格。

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。

board.length == board[i].length == 8board[i][j]可以是'r'，'.'，'b'或'p'只有乙個格仔上存在board[i][j] == 'r'

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，棋盤中有乙個「車」，問「車」向上下左右四個方向遍歷能吃到多少個「卒」。條件是：

不能出棋盤；

遇到「象」不通；

一旦吃到「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 }}
// 向左
for (int j = rookj; j >= 0; j--) 
if (board[rooki][j] == 'p') 
}// 向右
for (int j = rookj; j < board[rooki].length; j++) 
if (board[rooki][j] == 'p') 
}// 向上
for (int i = rooki; i >= 0; i--) 
if (board[i][rookj] == 'p') 
}// 向下
for (int i = rooki; i < board.length; i++) 
if (board[i][rookj] == 'p') 
}return count;}}

LeetCode 999 車的可用捕獲量

LeetCode999 車的可用捕獲量

LeetCode 999 車的可用捕獲量

999 車的可用捕獲量

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