999.Available Captures for Rook
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'
代码如下：
class Solution {
public:
int numRookCaptures(vector<vector<char>>& board) {
vector<int> R={0,0};
for(int i=0;i<8;i++)
for(int j=0;j<8;j++)
if(board[i][j]=='R')
{
R[0]=i;
R[1]=j;
break;
}
int cnt=0;
for(int k=R[1];k<8;k++)
if(board[R[0]][k]=='p'){cnt++;break;}
else if(board[R[0]][k]=='B') break;
for(int k=R[0];k<8;k++)
if(board[k][R[1]]=='p'){cnt++;break;}
else if(board[k][R[1]]=='B') break;
for(int k=R[1];k>=0;k--)
if(board[R[0]][k]=='p') {cnt++;break;}
else if(board[R[0]][k]=='B') {break;}
for(int k=R[0];k>=0;k--)
if(board[k][R[1]]=='p'){cnt++;break;}
else if(board[k][R[1]]=='B') break;
return cnt;
}
};
执行用时 : 8 ms, 在Available Captures for Rook的C++提交中击败了100.00% 的用户 内存消耗 : 8.4 MB, 在Available Captures for Rook的C++提交中击败了100.00% 的用户
