999. 车的可用捕获量

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  • 用户通过次数255
  • 用户尝试次数260
  • 通过次数255
  • 提交次数357
  • 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:

  1. board.length == board[i].length == 8
  2. board[i][j] 可以是 'R''.''B' 或 'p'
  3. 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

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