LeetCode——999.车的可用捕获量js

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本文介绍了一个算法，用于计算国际象棋中白色车在8x8棋盘上能够一次移动捕获到的黑色卒数量。通过分析棋盘布局，避免与友方象冲突，实现对卒的有效捕获。

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题目描述:

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。

示例1：
在这里插入图片描述

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：
在这里插入图片描述

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：
在这里插入图片描述

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。

题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/available-captures-for-rook

解法:

  var numRookCaptures = function (board) {
            var b = 0, Iindex = 0, Jindex = 0;
            //获取车的位置;
            top: for (var i = 0; i < board.length; i++) {
                for (var j = 0; j < board[i].length; j++) {
                    if (board[i][j] === "R") {
                        Iindex = i;
                        Jindex = j;
                        break top;
                    }
                }
            }
           /*判断车的上下左右距离其最近的棋子
             如果是象，不操作，跳出循坏，表示距离其最近的是象，则捕获不到卒。
            如果是卒，b++,此时跳出循环，表示一侧有多个卒时只能捕获到一个。*/
            for (var n = Jindex; n > 0; n--) {    //左
                if (board[Iindex][n] === "p") {
                    b++; break;
                }
                if (board[Iindex][n] === "B") {
                    break;
                }
            }
            for (var t = Jindex; t < 8; t++) {   //右
                if (board[Iindex][t] === "p") {
                    b++; 
                    break;
                }
                if (board[Iindex][t] === "B") {
                    break;
                } 
            }   
            for (var m = Iindex; m > 0; m--) { //上
                if (board[m][Jindex] === "p") {
                    b++; break;
                }
                if (board[m][Jindex] === "B") {
                    break;
                }
            }
            for (var o = Iindex; o < 8; o++) { //下
                if (board[o][Jindex] === "p") {
                    b++; break;
                }
                if (board[o][Jindex] === "B") {
                    break;
                }
            }
            return b;
        };