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'
思路：这还要啥思路，水题无疑。
1 class Solution { 2 public: 3 int numRookCaptures(vector<vector<char>>& board) { 4 int col= board.size(); 5 int row = board[0].size(); 6 int count = 0; 7 for(int i = 0;i<col;i++) 8 { 9 for(int j = 0;j<row;j++) 10 { 11 if(board[i][j] == 'R') 12 { 13 14 for(int m = j+1;m < row;m++) 15 { 16 //north 17 char c = board[i][m]; 18 if(c== 'p') 19 { 20 cout<<"north:"<<i<<","<<m<<endl; 21 count++; 22 break; 23 } 24 else if(c == 'B') 25 { 26 break; 27 } 28 } 29 for(int m = i+1;m<row;m++) 30 { 31 //east 32 char c = board[m][j]; 33 if(c== 'p') 34 { 35 cout<<"east:"<<m<<","<<j<<endl; 36 count++; 37 break; 38 } 39 else if(c == 'B') 40 { 41 break; 42 } 43 } 44 for(int m = i-1;m>=0;m--) 45 { 46 //west 47 char c = board[m][j]; 48 if(c== 'p') 49 { 50 cout<<"west:"<<m<<","<<j<<endl; 51 count++; 52 break; 53 } 54 else if(c == 'B') 55 { 56 break; 57 } 58 } 59 for(int m = j-1;m>=0;m--) 60 { 61 //south 62 char c = board[i][m]; 63 if(c== 'p') 64 { 65 cout<<"south:"<<i<<","<<m<<endl; 66 count++; 67 break; 68 } 69 else if(c == 'B') 70 { 71 break; 72 } 73 } 74 break; 75 } 76 } 77 } 78 return count; 79 } 80 };