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 

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