999.可以被一步捕获的棋子数

【LetMeFly】999.可以被一步捕获的棋子数：模拟

力扣题目链接：https://leetcode.cn/problems/available-captures-for-rook/

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。

 

示例 1：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够吃掉所有的卒。

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：
象阻止了车吃掉任何卒。

示例 3：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 
车可以吃掉位置 b5，d6 和 f5 的卒。

 

提示：

1. board.length == 8
2. board[i].length == 8
3. board[i][j] 可以是 'R'，'.'，'B' 或 'p'
4. 只有一个格子上存在 board[i][j] == 'R'

解题方法：模拟

一共分为两步：

1. 遍历棋盘，遇到字符R时停下，并记录下起点下标
2. 从起点开始分别向上下左右四个方向遍历，遇到边界或者遇到B停止。同时，遍历时若遇到p，则答案数量加一并停止。

• 时间复杂度$O(mn)$，其中棋盘大小为$m\times n$
• 空间复杂度$O(1)$

我的一个评论

AC代码

C++

const int directions[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int x, y;
        for (x = 0; x < board.size(); x++) {
            for (y = 0; y < board[0].size(); y++) {
                if (board[x][y] == 'R') {
                    goto loop;
                }
            }
        }
        loop:;
        int ans = 0;
        for (int d = 0; d < 4; d++) {
            for (int step = 1; ; step++) {
                int nx = x + directions[d][0] * step;
                int ny = y + directions[d][1] * step;
                if (nx < 0 || nx >= board.size() || ny < 0 || ny >= board[0].size() || board[nx][ny] == 'B') {
                    break;
                }
                if (board[nx][ny] == 'p') {
                    ans++;
                    break;
                }
            }
        }
        return ans;
    }
};

同步发文于CSDN和我的个人博客，原创不易，转载经作者同意后请附上原文链接哦~

Tisfy：https://letmefly.blog.csdn.net/article/details/144303138