1222. Queens That Can Attack the King**

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本文介绍了解决LeetCode上1222题QueensThatCanAttacktheKing的两种C++实现方法。问题要求找出所有能攻击到特定国王位置的黑色皇后。通过遍历八个方向寻找最近的皇后，或检查所有可能的格子并使用Hash表存储皇后位置来解决。

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;
    }
};