问题：
给定8*8棋盘中，queen的坐标，和king的坐标。
king的同一行，同一列，同一对角线上的第一个queen，为可攻击king的queen
求所有可攻击king的queen的坐标数组。
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[0].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.
解法：
一共有8个方向，可作为要求的queen
分别为：king 的右方→，左方←，下方↓，上方↑，左上↖️，右上↗️，左下↙️，右下↘️。
那么构建待选数组cand
vector<vector<int>> cand={{0,1},{0,-1},{1,0},{-1,0},{1,1},{-1,-1},{1,-1},{-1,1}};
将所有待选queens放入unordered_set中方便查找。
分别试探以king为起始，向8个方向逐次+1距离的节点，是否在queens里面。
在的话，加入res。
代码参考：
1 class Solution { 2 public: 3 vector<vector<int>> queensAttacktheKing(vector<vector<int>>& queens, vector<int>& king) { 4 vector<vector<int>> cand={{0,1},{0,-1},{1,0},{-1,0},{1,1},{-1,-1},{1,-1},{-1,1}}; 5 unordered_set<int> qs; 6 vector<vector<int>> res; 7 for(vector<int> queen:queens){ 8 qs.insert(queen[0]*8+queen[1]); 9 } 10 for(int i=0; i<8; i++){ 11 int x=king[0], y=king[1]; 12 while(x>=0 && x<8 && y>=0 && y<8){ 13 if(qs.find(x*8+y)!=qs.end()){ 14 res.push_back({x,y}); 15 break; 16 } 17 x+=cand[i][0]; 18 y+=cand[i][1]; 19 } 20 } 21 return res; 22 } 23 };