[LeetCode] 1222. Queens That Can Attack the King 可以攻擊國王的皇后

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.

這道題說是在一個 8 by 8 大小的國際象棋的棋盤上，放了多個黑色皇后和一個白色的國王，國際象棋的中的皇后就相當於中國象棋中的車，不過比車更叼，不但能橫豎走，還能斜着走，國王就相當於將和帥，現在問有多少個皇后能攻擊到國王。題目中的每個例子都十分貼心的配了圖，可以更好的幫助我們理解。博主最開始沒看清楚題目，認為直接判斷每個皇后是否跟國王共線就可以了，但是即便是共線，也不代表皇后就可以攻擊到國王，若中間還有其他皇后就不行，就像例子1中的那樣。所以視角就應該轉移到國王本身，對於棋盤上任意一個非邊緣位置，共有八個方向可以去，而每個方向都可能遇到皇后，一旦遇到皇后了，這個方向就不會再遇到其他的皇后了。所以就應該以國王為起點，分別朝八個方向前進，看是否可以遇到皇后。

博主之前寫出的解法有點復雜，是分別遍歷八個方向的情況，但實際上可以寫的更加簡潔一些，利用方向變量的 offset，就像在迷宮遍歷時，寫出四個方向的變量一樣，這里八個方向其實就是 [-1, 0, 1] 中任取兩個數字分別加上當前的橫縱坐標，注意要去掉 (0, 0) 這種情況，因為並不會發生位置的變化。所以當拿到一組偏移量時，就可以進行 while 循環，條件時不越界，當新位置上有皇后時，將該位置加入結果 res，並 break 掉循環，否則再次加上偏移量繼續循環。如何快速的知道某個位置上是否有皇后呢，當然不能每次遍歷一次所有皇后的位置，太不高效，這里可以建立另一個 8 by 8 大小的二維數組 seen，當 seen[i][j] 為1時，就表示 (i, j) 位置是皇后，參見代碼如下：

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

參考資料：

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

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