Leetcode 999. Available Captures for Rook , Java解法。

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博客围绕8x8棋盘展开，棋盘上有白车、空格、白象和黑卒，分别用字符 'R'、'.'、'B' 和 'p' 表示。介绍了白车的移动规则，需计算白车一步可捕获的黑卒数量，还提及时间复杂度为O（N^2）。

On an 8 x 8 chessboard, there is one white rook. There also may be empty squares, white bishops, and black pawns. These are given as characters 'R', '.', 'B', and 'p' respectively. Uppercase characters represent white pieces, and lowercase characters represent black pieces.

The rook moves as in the rules of Chess: it chooses one of four cardinal directions (north, east, west, and south), then moves in that direction until it chooses to stop, reaches the edge of the board, or captures an opposite colored pawn by moving to the same square it occupies. Also, rooks cannot move into the same square as other friendly bishops.

Return the number of pawns the rook can capture in one move.

Example 1:

Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation: 
In this example the rook is able to capture all the pawns.

Approach #1

class Solution {
        public int numRookCaptures(char[][] board) {
        int x0 = 0, y0 = 0, res = 0;
        // first find the Rook
        for (int i = 0; i < 8; i++)
            for (int j = 0; j < 8; j++)
                if (board[i][j] == 'R') {
                    x0 = i;
                    y0 = j;
                }
         
         // check each direction
        for (int[] d : new int[][] {{1, 0}, {0, 1}, { -1, 0}, {0, -1}}) {
            for (int x = x0 + d[0], y = y0 + d[1];
                 0 <= x && x < 8 && 0 <= y && y < 8;
                 x += d[0], y += d[1]) {
                
                    if (board[x][y] == 'p') 
                        res++;
                    if (board[x][y] != '.') 
                        break;
            }
        }
        return res;
    }
}

时间复杂度 O（N^2）

借鉴一下

for (int[] d : new int[][] {{1, 0}, {0, 1}, { -1, 0}, {0, -1}})

这一行的优雅。