[LeetCode] 999. Available Captures for Rook

You are given an 8 x 8 matrix representing a chessboard. There is exactly one white rook represented by 'R', some number of white bishops 'B', and some number of black pawns 'p'. Empty squares are represented by '.'.

A rook can move any number of squares horizontally or vertically (up, down, left, right) until it reaches another piece or the edge of the board. A rook is attacking a pawn if it can move to the pawn's square in one move.

Note: A rook cannot move through other pieces, such as bishops or pawns. This means a rook cannot attack a pawn if there is another piece blocking the path.

Return the number of pawns the white rook is attacking.

Example 1:

Input: board = [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3

Explanation:
In this example, the rook is attacking all the pawns.

Example 2:

Input: board = [[".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 0

Explanation:
The bishops are blocking the rook from attacking any of the pawns.

Example 3:

Input: board = [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3

Explanation:
The rook is attacking the pawns at positions b5, d6, and f5.

Constraints:
board.length == 8
board[i].length == 8
board[i][j] is either 'R', '.', 'B', or 'p'
There is exactly one cell with board[i][j] == 'R'

可以被一步捕获的棋子数。

给定一个 8 x 8 的棋盘，只有一个 白色的车，用字符 'R' 表示。棋盘上还可能存在白色的象 'B' 以及黑色的卒 'p'。空方块用字符 '.' 表示。

车可以按水平或竖直方向（上，下，左，右）移动任意个方格直到它遇到另一个棋子或棋盘的边界。如果它能够在一次移动中移动到棋子的方格，则能够 吃掉 棋子。

注意：车不能穿过其它棋子，比如象和卒。这意味着如果有其它棋子挡住了路径，车就不能够吃掉棋子。

返回白车将能 吃掉 的 卒的数量。

思路

这个题没有什么算法，就是按规则从车所在的位置开始扫描 4 个不同的方向，遇到小写的 p 则 res++，遇到边缘和大写的字母（因为是白色棋子）则停下。这个题就是考你在矩阵中如何一直往一个方向走的技巧。

复杂度

时间O(64) - 最极端情况需要扫描整个棋盘
空间O(1)

代码

Java实现

class Solution {
    public int numRookCaptures(char[][] board) {
        int m = board.length;
        int n = board[0].length;
        int[][] dirs = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };
        int res = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                // when find the rook
                if (board[i][j] == 'R') {
                    for (int[] dir : dirs) {
                        int x = i;
                        int y = j;
                        while (true) {
                            x += dir[0];
                            y += dir[1];
                            if (x < 0 || y < 0 || x >= m || y >= n || board[x][y] == 'B') {
                                break;
                            }
                            if (board[x][y] == 'p') {
                                res++;
                                break;
                            }
                        }
                    }
                }
            }
        }
        return res;
    }
}