999. 车的可用捕获量

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'

解法一

//时间复杂度O(1), 空间复杂度O(1)
class Solution {
public:int numRookCaptures(vector<vector<char>>& board, int i, int j) {int count = 0, k;for(k = i + 1; k < 8 && board[k][j] == '.'; k++);if(k != 8 && board[k][j] == 'p') count++;for(k = i - 1; k >= 0 && board[k][j] == '.'; k--);if(k != -1 && board[k][j] == 'p') count++;for(k = j + 1; k < 8 && board[i][k] == '.'; k++);if(k != 8 && board[i][k] == 'p') count++;for(k = j - 1; k >= 0 && board[i][k] == '.'; k--);if(k != -1 && board[i][k] == 'p') count++;return count;}int numRookCaptures(vector<vector<char>>& board) {for(int i = 0; i < 8; i++) {for(int j = 0; j < 8; j++) {if(board[i][j] == 'R') return numRookCaptures(board, i, j);}}return 0;}
};

思路：

先遍历一次寻找车（‘R’），找到之后从其位置出发，向上、下、左、右4个方向寻找，若找到卒（‘p’）就将计数count++，最后返回count。