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999.Available Captures for Rook
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：
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输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
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输出：3
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解释：
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在本例中，车能够捕获所有的卒。
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示例 2： 
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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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
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输出：0
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解释：
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象阻止了车捕获任何卒。
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示例 3：
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输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
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输出：3
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解释：
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车可以捕获位置 b5，d6 和 f5 的卒。
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提示：
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board.length == board[i].length == 8
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board[i][j] 可以是 'R'，'.'，'B' 或 'p'
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只有一个格子上存在 board[i][j] == 'R'
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代码如下：
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class Solution {
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public:
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int numRookCaptures(vector<vector<char>>& board) {
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vector<int> R={0,0};
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for(int i=0;i<8;i++)
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for(int j=0;j<8;j++)
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if(board[i][j]=='R')
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{
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R[0]=i;
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R[1]=j;
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break;
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}
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int cnt=0;
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for(int k=R[1];k<8;k++)
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if(board[R[0]][k]=='p'){cnt++;break;}
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else if(board[R[0]][k]=='B') break;
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for(int k=R[0];k<8;k++)
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if(board[k][R[1]]=='p'){cnt++;break;}
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else if(board[k][R[1]]=='B') break;
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for(int k=R[1];k>=0;k--)
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if(board[R[0]][k]=='p') {cnt++;break;}
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else if(board[R[0]][k]=='B') {break;}
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for(int k=R[0];k>=0;k--)
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if(board[k][R[1]]=='p'){cnt++;break;}
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else if(board[k][R[1]]=='B') break;
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return cnt;
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}
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};
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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
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