1352. Product of the Last K Numbers

On an 8x8 chessboard, there can be multiple Black Queens and one White King.

Given an array of integer coordinates queens that represents the positions of the Black Queens, and a pair of coordinates king that represent the position of the White King, return the coordinates of all the queens (in any order) that can attack the King.

Example 1:

Input: queens = [[0,1],[1,0],[4,0],[0,4],[3,3],[2,4]], king = [0,0]
Output: [[0,1],[1,0],[3,3]]
Explanation: 
The queen at [0,1] can attack the king cause they're in the same row.
The queen at [1,0] can attack the king cause they're in the same column.
The queen at [3,3] can attack the king cause they're in the same diagnal.
The queen at [0,4] can't attack the king cause it's blocked by the queen at [0,1].
The queen at [4,0] can't attack the king cause it's blocked by the queen at [1,0].
The queen at [2,4] can't attack the king cause it's not in the same row/column/diagnal as the king.

Example 2:

Input: queens = [[0,0],[1,1],[2,2],[3,4],[3,5],[4,4],[4,5]], king = [3,3]
Output: [[2,2],[3,4],[4,4]]

Example 3:

Input: queens = [[5,6],[7,7],[2,1],[0,7],[1,6],[5,1],[3,7],[0,3],[4,0],[1,2],[6,3],[5,0],[0,4],[2,2],[1,1],[6,4],[5,4],[0,0],[2,6],[4,5],[5,2],[1,4],[7,5],[2,3],[0,5],[4,2],[1,0],[2,7],[0,1],[4,6],[6,1],[0,6],[4,3],[1,7]], king = [3,4]
Output: [[2,3],[1,4],[1,6],[3,7],[4,3],[5,4],[4,5]]

Constraints:

• 1 <= queens.length <= 63
• queens[i].length == 2
• 0 <= queens[i][j] < 8
• king.length == 2
• 0 <= king[0], king[1] < 8
• At most one piece is allowed in a cell.

这道题说是在一个 8 by 8 大小的国际象棋的棋盘上，放了多个黑色皇后和一个白色的国王，国际象棋的中的皇后就相当于中国象棋中的车，不过比车更叼，不但能横竖走，还能斜着走，国王就相当于将和帅，现在问有多少个皇后能攻击到国王。题目中的每个例子都十分贴心的配了图，可以更好的帮助我们理解。博主最开始没看清楚题目，认为直接判断每个皇后是否跟国王共线就可以了，但是即便是共线，也不代表皇后就可以攻击到国王，若中间还有其他皇后就不行，就像例子1中的那样。所以视角就应该转移到国王本身，对于棋盘上任意一个非边缘位置，共有八个方向可以去，而每个方向都可能遇到皇后，一旦遇到皇后了，这个方向就不会再遇到其他的皇后了。所以就应该以国王为起点，分别朝八个方向前进，看是否可以遇到皇后。

博主之前写出的解法有点复杂，是分别遍历八个方向的情况，但实际上可以写的更加简洁一些，利用方向变量的 offset，就像在迷宫遍历时，写出四个方向的变量一样，这里八个方向其实就是 [-1, 0, 1] 中任取两个数字分别加上当前的横纵坐标，注意要去掉 (0, 0) 这种情况，因为并不会发生位置的变化。所以当拿到一组偏移量时，就可以进行 while 循环，条件时不越界，当新位置上有皇后时，将该位置加入结果 res，并 break 掉循环，否则再次加上偏移量继续循环。如何快速的知道某个位置上是否有皇后呢，当然不能每次遍历一次所有皇后的位置，太不高效，这里可以建立另一个 8 by 8 大小的二维数组 seen，当 seen[i][j] 为1时，就表示 (i, j) 位置是皇后，参见代码如下：

class Solution {
public:
    vector<vector<int>> queensAttacktheKing(vector<vector<int>>& queens, vector<int>& king) {
        vector<vector<int>> res, seen(8, vector<int>(8));
        for (auto queen : queens) {
            seen[queen[0]][queen[1]] = 1;
        }
        for (int i = -1; i <= 1; ++i) {
            for (int j = -1; j <= 1; ++j) {
                if (i == 0 && j == 0) continue;
                int x = king[0] + i, y = king[1] + j;
                while (min(x, y) >= 0 && max(x, y) < 8) {
                    if (seen[x][y] == 1) {
                        res.push_back({x, y});
                        break;
                    }
                    x += i, y += j;
                }
            }
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1222

参考资料：

https://leetcode.com/problems/queens-that-can-attack-the-king/

https://leetcode.com/problems/queens-that-can-attack-the-king/discuss/403755/C%2B%2B-Tracing

https://leetcode.com/problems/queens-that-can-attack-the-king/discuss/403687/Java-short-and-concise-beat-100

LeetCode All in One 题目讲解汇总(持续更新中…)

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