题目
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
每次可以在四个方向上分别移动一次，先找出'R'位置，然后在四个方向上移动。
代码
时间复杂度：O(n^2)
空间复杂度：O(1)
class Solution {
public:
int numRookCaptures(vector<vector<char>>& board) {
vector<int> dx = {0, 1, 0, -1}, dy = {1, 0, -1, 0};
int tx, ty, res = 0;
//1. 找出R位置
for (int i = 0; i < 8; ++i) {
for (int j = 0; j < 8; ++j) {
if (board[i][j] == 'R') {
tx = i;
ty = j;
break;
}
}
}
//2. 分别朝四个方向上移动
for (int i = 0; i < 4; ++i) {
int step = 0;
while (step < 8) {
int x = tx + step * dx[i];
int y = ty + step * dy[i];
if (x < 0 || x >= 8 || y < 0 || y >= 8 || board[x][y] == 'B') break;
if (board[x][y] == 'p') {
++res;
break;
}
++step;
}
}
return res;
}
};