LeetCode: Spiral Matrix II

一次过

class Solution {
public:
    bool out(int x, int y, int n) {
        if (x >= n || x < 0 || y >= n || y < 0) return true;
            return false;
    }
    void dfs(int x, int y, vector<vector<int>> &ret, int dir[4][2], int dep, int n, int direct) {
        if (dep == n*n) return;
        ret[x][y] = dep+1;
        if (out(x+dir[direct][0], y+dir[direct][1], n) || ret[x+dir[direct][0]][y+dir[direct][1]])
            direct = (direct+1)%4;
        dfs(x+dir[direct][0], y+dir[direct][1], ret, dir, dep+1, n, direct);
    }
    vector<vector<int> > generateMatrix(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> ret(n, vector<int>(n, 0));
        if (!n) return ret;
        int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        dfs(0, 0, ret, dir, 0, n, 0);
        return ret;
    }
};

 这段是iterative的方法，推荐

class Solution {
public:
    vector<vector<int> > generateMatrix(int n) {
        int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        vector<vector<int> > ans(n, vector<int>(n));
        int x = 0;
        int y = 0;
        int cur = 1;
        for (int i = 0; i < n/2; ++i) {
            for (int j = 0; j < 4; ++j) {
                for (int k = 0; k < n-1-2*i; ++k) {
                    ans[x][y] = cur++;
                    x += dir[j][0];
                    y += dir[j][1];
                }
            }
            x += 1;
            y += 1;
        }
        if (n % 2) ans[n/2][n/2] = cur;
        return ans;
    }
};

 C#

public class Solution {
    public int[,] GenerateMatrix(int n) {
        int[,] dir = new int[4, 2] {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int[,] ans = new int[n, n];
        int x = 0, y = 0, cur = 1;
        for (int i = 0; i < n/2; i++) {
            for (int j = 0; j < 4; j++) {
                for (int k = 0; k < n-1-2*i; k++) {
                    ans[x, y] = cur++;
                    x += dir[j, 0];
                    y += dir[j, 1];
                }
            }
            x++;
            y++;
        }
        if (n % 2 == 1) ans[n/2, n/2] = cur;
        return ans;
    }
}