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在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。
车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：
你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。
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 == 8board[i][j] 可以是 'R'，'.'，'B' 或 'p'
board[i][j] == 'R'
先找到 R 的位置，之后向“上、下、左、右”四个方向查找，累加结果。
class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
x, y, n = 0, 0, 8
for i in range(n):
for j in range(n):
if board[i][j] == 'R':
x, y = i, j
break
ans = 0
for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
i, j = x, y
while 0 <= i + a < n and 0 <= j + b < n and board[i + a][j + b] != 'B':
i, j = i + a, j + b
if board[i][j] == 'p':
ans += 1
break
return ans
class Solution {
public int numRookCaptures(char[][] board) {
int[] pos = find(board);
int ans = 0, n = 8;
int[][] dirs = new int[][] {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
for (int[] dir : dirs) {
int x = pos[0], y = pos[1], a = dir[0], b = dir[1];
while (
x + a >= 0 && x + a < n && y + b >= 0 && y + b < n && board[x + a][y + b] != 'B') {
x += a;
y += b;
if (board[x][y] == 'p') {
++ans;
break;
}
}
}
return ans;
}
private int[] find(char[][] board) {
int n = 8;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == 'R') {
return new int[] {i, j};
}
}
}
return null;
}
}
class Solution {
public:
int numRookCaptures(vector<vector<char>>& board) {
vector<int> pos = find(board);
int ans = 0, n = 8;
vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
for (auto& dir : dirs) {
int x = pos[0], y = pos[1], a = dir[0], b = dir[1];
while (x + a >= 0 && x + a < n && y + b >= 0 && y + b < n && board[x + a][y + b] != 'B') {
x += a;
y += b;
if (board[x][y] == 'p') {
++ans;
break;
}
}
}
return ans;
}
vector<int> find(vector<vector<char>>& board) {
int n = 8;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == 'R') {
return {i, j};
}
}
}
return {};
}
};
func numRookCaptures(board [][]byte) int {
n := 8
find := func() []int {
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
if board[i][j] == 'R' {
return []int{i, j}
}
}
}
return []int{}
}
pos := find()
ans := 0
dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
for _, dir := range dirs {
x, y, a, b := pos[0], pos[1], dir[0], dir[1]
for x+a >= 0 && x+a < n && y+b >= 0 && y+b < n && board[x+a][y+b] != 'B' {
x += a
y += b
if board[x][y] == 'p' {
ans++
break
}
}
}
return ans
}
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