leetcode 889. Spiral Matrix III

On a 2 dimensional grid with R rows and C columns, we start at (r0, c0) facing east.

Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column.

Now, we walk in a clockwise spiral shape to visit every position in this grid.

Whenever we would move outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.)

Eventually, we reach all R * C spaces of the grid.

Return a list of coordinates representing the positions of the grid in the order they were visited.

Example 1:

Input: R = 1, C = 4, r0 = 0, c0 = 0
Output: [[0,0],[0,1],[0,2],[0,3]]

Example 2:

Input: R = 5, C = 6, r0 = 1, c0 = 4
Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]

思路：8个方向去找，比如第一个位置1，找他的8个方向，第二个位置2，也去找他的8个方向，但是8个方向的查找次序要从(1->2)这个方向开始，这样能保证顺时针的顺序。

class Solution {
public:
    int dir[16][2] = {0,1,1,1,1,0,1,-1,0,-1,-1,-1,-1,0,-1,1,0,1,1,1,1,0,1,-1,0,-1,-1,-1,-1,0,-1,1};
    //int dir2[8][2] = {0,-1,-1,-1,-1,0,-1,1,0,1,1,1,1,0,1,-1};
    vector<vector<int> > spiralMatrixIII(int R, int C, int r0, int c0) {
        vector<vector<int> > ans;
        queue<pair<int,int> >q;
        map<pair<int,int>, int> mp;
        ans.push_back({r0, c0});
        mp[{r0, c0}] = 1;
        map<pair<int,int>, pair<int,int> > mp2;
        for (int i = 0; i < 8; ++i) {
            int x = r0 + dir[i][0];
            int y = c0 + dir[i][1];
            if (!mp[{x,y}] && x < R && y < C && x >= 0 && y >= 0) {
                mp2[{x,y}] = {dir[i][0],dir[i][1]};
                q.push({x,y});
                mp[{x,y}] = 1;
            }
        }
        while (!q.empty()) {
            pair<int,int> u = q.front(); q.pop();
            ans.push_back({u.first, u.second});
            int i = 0;
            pair<int,int> p = mp2[{u.first,u.second}];
            int ox = -p.first;
            int oy = -p.second;
             //cout << "NO";
            //cout << ox << " " << oy << endl;
            int mark  = 0;
            while (i < 16) {
                if (dir[i][0] == ox && dir[i][1] == oy) {
                    mark = i;
                    break;
                }
                ++i;
            }
            for (; i < mark+8; ++i) {
                //cout << i << endl;
                int x = u.first + dir[i][0];
                int y = u.second + dir[i][1];
                if (!mp[{x,y}] && x < R && y < C && x >= 0 && y >= 0) {
                    q.push({x,y});
                    mp2[{x,y}] = {dir[i][0],dir[i][1]};
                    mp[{x,y}] = 1;
                }
            }
        }
        return ans;
    }
};