Leetcode 遍歷模擬題目螺旋矩陣

模擬題目，使用額外的空間來記錄是否遍歷過，時間複雜度O(n^2)

const int dx[]={1,0,-1,0};
const int dy[]={0,1,0,-1};
class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        vector<int> res;
        int x = 0, y = 0, i = 0, n = matrix.size();
        if(n==0) return res;
        int m = matrix[0].size();
        vector<vector<bool>> isVisited(n,vector<bool>(m,false));
        isVisited[0][0]=true;
        res.push_back(matrix[x][y]);
        for(int j=1;j<n*m;j++){
            //cout<<j<<endl;
            int a = x + dx[i];
            int b = y + dy[i];
            //cout<<j<<" "<<a<<endl;
            if(a>=m||a<0||b>=n||b<0||isVisited[b][a]){
                i=(i+1)%4;
                a = x + dx[i];
                b = y + dy[i];
            }
            x = a;
            y = b;
            isVisited[b][a] = true;
            res.push_back(matrix[b][a]);
        }
        return res;
    }
};

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        const int dx[]={1,0,-1,0};
        const int dy[]={0,1,0,-1};
        int x = 0, y = 0, i = 0;
        if(n==0) return {};
        vector<vector<int>> res(n,vector<int>(n,0));
        res[0][0] = 1;
        for(int j=2;j<=n*n;j++){
            int a = x + dx[i];
            int b = y + dy[i];
            if(a>=n||a<0||b>=n||b<0||res[b][a]){
                i=(i+1)%4;
                a = x + dx[i];
                b = y + dy[i];
            }
            x = a;
            y = b;
            res[b][a] = j;
        }
        return res;
    }
};