day13 (999. 车的可用捕获量)

• 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’

• 我的题解：

  class Solution {
      public int numRookCaptures(char[][] board) {
          int res = 0;
          for(int i = 0; i < 8;i++) {
              for(int j = 0; j < 8; j++) {
                  if(board[i][j] == 'R') {
                      int p = i,q=j;
                      while(p < 8) {
                          if(board[p][q] == 'B') {
                              break;
                          }else if(board[p][q] == 'p') {
                              res++;
                              break;
                          }
                          p++;
                      }
                      p = i;
                      while(p >= 0) {
                          if(board[p][q] == 'B') {
                              break;
                          }else if(board[p][q] == 'p') {
                              res++;
                              break;
                          }
                          p--;
                      }
                      p = i;
                      while(q < 8) {
                          if(board[p][q] == 'B') {
                              break;
                          }else if(board[p][q] == 'p') {
                              res++;
                              break;
                          }
                          q++;
                      }
                      q = j;
                      while(q >= 0) {
                          if(board[p][q] == 'B') {
                              break;
                          }else if(board[p][q] == 'p') {
                              res++;
                              break;
                          }
                          q--;
                      }
                      break;
                  }
              }
          }
          return res;
      }
  }
  

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