1222. Queens That Can Attack the King**

1222. Queens That Can Attack the King**

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

题目描述

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.

Example1:

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.

还有几个例子见原网页吧.

C++ 实现 1

向 8 个方向分别搜索最近的 ​​Queen​​. 思路来自 LeetCode 的 Submission.

class Solution {
private:
    void findQueensByDirection(const vector<vector<int>>& queens,
                              const vector<int>& king,
                              vector<vector<int>>& res,
                              int i, int j) {
        int x = king[0], y = king[1];
        while (x >= 0 && x < 8 && y >= 0 && y < 8) {
            x += i;
            y += j;
            if (std::find(queens.begin(),
                          queens.end(),
                          vector<int>{x, y}) != queens.end()) {
                res.push_back({x, y});
                return;
            }
        }
    }
public:
    vector<vector<int>> queensAttacktheKing(vector<vector<int>>& queens, vector<int>& king) {
        vector<vector<int>> res;
        for (auto &i : {-1, 0, 1}) {
            for (auto &j : {-1, 0, 1}) {
                if (i == 0 && j == 0) continue;
                findQueensByDirection(queens, king, res, i, j);
            }
        }
        return res;
    }
};

C++ 实现 2

思路来自: ​​[Python] Check 8 steps in 8 Directions​​​, 检查 8 个方向的所有格子, 使用 Hash 表保存 ​​Queens​​ 的坐标, 判断 queen 是否在 Hash 表中.

class Solution {
private:
    // https://stackoverflow.com/questions/29855908/c-unordered-set-of-vectors
    struct VectorHash {
        size_t operator()(const std::vector<int>& v) const {
            std::hash<int> hasher;
            size_t seed = 0;
            for (int i : v) {
                seed ^= hasher(i) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
            }
            return seed;
        }
    };
public:
    vector<vector<int>> queensAttacktheKing(vector<vector<int>>& queens, vector<int>& king) {
        unordered_set<vector<int>, VectorHash> records;
        vector<vector<int>> res;
        for (auto &p : queens) records.insert(p);
        for (auto &i : {-1, 0, 1}) {
            for (auto &j : {-1, 0, 1}) {
                for (auto &k : {1, 2, 3, 4, 5, 6, 7}) {
                    int qx = king[0] + i * k, qy = king[1] + j * k;
                    if (records.count({qx, qy})) {
                        res.push_back({qx, qy});
                        break;
                    }
                }
            }
        }
        return res;
    }
};