day13 (999. 車的可用捕獲量)

• 題目描述：在一個 8 x 8 的棋盤上，有一個白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字符 “R”，“.”，“B” 和 “p” 給出。大寫字符表示白棋，小寫字符表示黑棋。

  車按國際象棋中的規則移動：它選擇四個基本方向中的一個（北，東，西和南），然後朝那個方向移動，直到它選擇停止、到達棋盤的邊緣或移動到同一方格來捕獲該方格上顏色相反的卒。另外，車不能與其他友方（白色）象進入同一個方格。

  返回車能夠在一次移動中捕獲到的卒的數量。

• 示例 1：
  

  輸入：[[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",“R”,".",".",".",“p”],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
  輸出：3
  解釋：
  在本例中，車能夠捕獲所有的卒。

• 示例 2：
  

  輸入：[[".",".",".",".",".",".",".","."],[".",“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”,".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
  輸出：0
  解釋：
  象阻止了車捕獲任何卒。

• 示例 3：
  

  輸入：[[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",“p”,".",".",".","."],[“p”,“p”,".",“R”,".",“p”,“B”,"."],[".",".",".",".",".",".",".","."],[".",".",".",“B”,".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",".",".",".",".","."]]
  輸出：3
  解釋：
  車可以捕獲位置 b5，d6 和 f5 的卒。

• 提示：

  board.length == board[i].length == 8
  board[i][j] 可以是 ‘R’，’.’，‘B’ 或 ‘p’
  只有一個格子上存在 board[i][j] == ‘R’

• 我的題解：

  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;
      }
  }
  

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