go语言LeetCode题解999可以被一步捕获的棋子数
题目描述
999. 可以被一步捕获的棋子数 - 力扣(LeetCode)
在一个 8 x 8
的棋盘上,有一个白色的车(Rook
),用字符 'R'
表示。棋盘上还可能存在空方块,白色的象(Bishop
)以及黑色的卒(pawn
),分别用字符 '.'
,'B'
和 'p'
表示。不难看出,大写字符表示的是白棋,小写字符表示的是黑棋。
车按国际象棋中的规则移动。东,西,南,北四个基本方向任选其一,然后一直向选定的方向移动,直到满足下列四个条件之一:
- 棋手选择主动停下来。
- 棋子因到达棋盘的边缘而停下。
- 棋子移动到某一方格来捕获位于该方格上敌方(黑色)的卒,停在该方格内。
- 车不能进入/越过已经放有其他友方棋子(白色的象)的方格,停在友方棋子前。
你现在可以控制车移动一次,请你统计有多少敌方的卒处于你的捕获范围内(即,可以被一步捕获的棋子数)。
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。
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'
思路分析
这道题首先要理解题意
- 如果没有阻挡,车可以无限移动,除非自己停止
- 遇到象,停止。停止的意思是不能向前,但可以向后
- 遇到边,停止。停止的意思是不能向前,但可以向后
- 遇到卒,吃掉,然后在这个方向上必须停止。
解题方法,先整理
- 去掉没用的信息
- 把二维问题转为一维问题。
AC 代码
/** * @param {character[][]} board * @return {number} */ var numRookCaptures = function (board) { let count = 0 let info = [] for (let i = 0; i < 8; i++) { let item = [] for (let j = 0; j < 8; j++) { if ('.' !== board[i][j]) { item.push(board[i][j]) } } item.length > 0 && info.push(item) } for (let j = 0; j < 8; j++) { let item = [] for (let i = 0; i < 8; i++) { if ('.' !== board[i][j]) { item.push(board[i][j]) } } item.length > 0 && info.push(item) } //整理好后的info是个一维数组 for (let item of info) { let index = item.indexOf('R') if (index < 0) continue let i = index while (i--) { if (item[i] === 'B') break; if (item[i] === 'p') { count++ break } } i = index + 1 while (i < item.length) { if (item[i] === 'B') break; if (item[i] === 'p') { count++ break } i++ } } return count };
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