刷题55—车的可用捕获量

92.车的可用捕获量

题目链接

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

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：

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'

重难点

理解题意：

1. 首先找到'R';
2. 车可以移动的方向为：北、南、西、东；
3. 车选则一个方向移动，直到遇到边界或者'B'或者'p'停下来，从车的起始位置开始移动；
4. 因此车在移动过程中需要判断边界（回到初始位置）、'B'（回到初始位置）、'p'（回到初始位置，捕获卒的数量加1）。

题目分析

1. 首先找到'R';
2. 接下来根据车的移动方向遍历：北、南、西、东，以x轴正方向为东，以y轴正方向为北，四个方向可以用个二维数组{{0,1}，{0，-1}，{-1,0}，{1,0}}来表示；
3. 遍历棋盘的每个格子，找到'R'后，记录'R'当前的位置，分别将'R'从四个方向移动；
4. 若是遇到边界，就回到初始位置）；遇到'B'，就回到初始位置；遇到'p'就回到初始位置且捕获卒的数量加1；

/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function(board) {
    let num = 0;
    let n = board.length;
    for(let i=0;i<n;i++){
        for(let j=0;j<n;j++){
            if(board[i][j] === 'R'){
                let m = i;
                let n = j;
                //i-1 西
                while(--m >= 0){
                    if(board[m][n] === 'B') break;
                    if(board[m][n] === 'p'){
                        num++;
                        break;
                    } 
                }
                m = i;
                //i+1 东
                while(++m < 8){
                    if(board[m][n] === 'B') break;
                    if(board[m][n] === 'p'){
                        num++;
                        break;
                    } 
                }
                m = i;
                //j+1 北
                while(++n < 8){
                    if(board[m][n] === 'B') break;
                    if(board[m][n] === 'p'){
                        num++;
                        break;
                    } 
                }
                n = j;
                //j-1 南
                while(--n >= 0){
                    if(board[m][n] === 'B') break;
                    if(board[m][n] === 'p'){
                        num++;
                        break;
                    } 
                }
                return num;
            }
        }
    }
};