30 Day Challenge Day 21 | Leetcode 1197. Minimum Knight Moves

class Solution {
public:
    int minKnightMoves(int x, int y) {
        int steps = 0;
        
        unordered_set<string> s;
        
        queue<pair<int, int>> q;
        q.push({0, 0});
        
        s.insert(to_string(0) + '-' + to_string(0));
        
        int dirs[8][2] = {{-2, 1},
                          {-2, -1},
                          {2, 1},
                          {2, -1},
                          {-1, 2},
                          {-1, -2},
                          {1, 2},
                          {1, -2}
                         };
        
        while(!q.empty()) {
            int sz = q.size();
            for(int i = 0; i < sz; i++) {
                auto t = q.front();
                q.pop();
                
                if(t.first == x && t.second == y) return steps;
                
                for(int k = 0; k < 8; k++) {
                    int next_x = t.first + dirs[k][0];
                    int next_y = t.second + dirs[k][1];
                    if(!s.count(to_string(next_x) + '-' + to_string(next_y))) {
                        q.push({next_x, next_y});
                        s.insert(to_string(next_x) + '-' + to_string(next_y));
                    }
                }
            }
            steps++;
        }
        
        return -1;
    }
};

改进：

利用对称性：四个象限的坐标是具有无符号对称性的。那么不考虑象限，或者把所有坐标转换到第一象限，是不影响结果的。

class Solution {
public:
    int minKnightMoves(int x, int y) {
        // reduce the search space
        x = abs(x);
        y = abs(y);
        
        int steps = 0;
        
        unordered_set<string> s;
        
        queue<pair<int, int>> q;
        q.push({0, 0});
        
        s.insert(to_string(0) + '-' + to_string(0));
        
        int dirs[8][2] = {{-2, 1},
                          {-2, -1},
                          {2, 1},
                          {2, -1},
                          {-1, 2},
                          {-1, -2},
                          {1, 2},
                          {1, -2}
                         };
        
        while(!q.empty()) {
            int sz = q.size();
            for(int i = 0; i < sz; i++) {
                auto t = q.front();
                q.pop();
                
                if(t.first == x && t.second == y) return steps;
                
                for(int k = 0; k < 8; k++) {
                    int next_x = t.first + dirs[k][0];
                    int next_y = t.second + dirs[k][1];
                    if(!s.count(to_string(next_x) + '-' + to_string(next_y))
                      && next_x >= -2 && next_y >= -2 && next_x < x+2 && next_y < y+2) { // reduce the search space
                        q.push({next_x, next_y});
                        s.insert(to_string(next_x) + '-' + to_string(next_y));
                    }
                }
            }
            steps++;
        }
        
        return -1;
    }
};