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- 题目难度Easy
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'
class Solution {
public:
int numRookCaptures(vector<vector<char>>& board) {
int n=,m=;
for(int i=;i < board.size();i++){
for(int j=;j < board[].size();j++){
if(board[i][j] == 'R'){n=i;m=j;}
}
}
int res = ;
for(int i=n-;i>=;i--){
if(board[i][m] == 'B')break;
else if(board[i][m] == 'p'){res++;break;}
}
for(int i=n+;i<board.size();i++){
if(board[i][m] == 'B')break;
else if(board[i][m] == 'p'){res++;break;}
}
for(int j=m-;j>=;j--){
if(board[n][j] == 'B')break;
else if(board[n][j] == 'p'){res++;break;}
}
for(int j=m+;j<board[].size();j++){
if(board[n][j] == 'B')break;
else if(board[n][j] == 'p'){res++;break;}
}
return res;
}
};
-HAOSHUIA