LeetCode: Spiral Matrix II

一次过

 1 class Solution {
 2 public:
 3     bool out(int x, int y, int n) {
 4         if (x >= n || x < 0 || y >= n || y < 0) return true;
 5             return false;
 6     }
 7     void dfs(int x, int y, vector<vector<int>> &ret, int dir[4][2], int dep, int n, int direct) {
 8         if (dep == n*n) return;
 9         ret[x][y] = dep+1;
10         if (out(x+dir[direct][0], y+dir[direct][1], n) || ret[x+dir[direct][0]][y+dir[direct][1]])
11             direct = (direct+1)%4;
12         dfs(x+dir[direct][0], y+dir[direct][1], ret, dir, dep+1, n, direct);
13     }
14     vector<vector<int> > generateMatrix(int n) {
15         // Start typing your C/C++ solution below
16         // DO NOT write int main() function
17         vector<vector<int>> ret(n, vector<int>(n, 0));
18         if (!n) return ret;
19         int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
20         dfs(0, 0, ret, dir, 0, n, 0);
21         return ret;
22     }
23 };

这段是iterative的方法，推荐

 1 class Solution {
 2 public:
 3     vector<vector<int> > generateMatrix(int n) {
 4         int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
 5         vector<vector<int> > ans(n, vector<int>(n));
 6         int x = 0;
 7         int y = 0;
 8         int cur = 1;
 9         for (int i = 0; i < n/2; ++i) {
10             for (int j = 0; j < 4; ++j) {
11                 for (int k = 0; k < n-1-2*i; ++k) {
12                     ans[x][y] = cur++;
13                     x += dir[j][0];
14                     y += dir[j][1];
15                 }
16             }
17             x += 1;
18             y += 1;
19         }
20         if (n % 2) ans[n/2][n/2] = cur;
21         return ans;
22     }
23 };

 1 public class Solution {
 2     public int[,] GenerateMatrix(int n) {
 3         int[,] dir = new int[4, 2] {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
 4         int[,] ans = new int[n, n];
 5         int x = 0, y = 0, cur = 1;
 6         for (int i = 0; i < n/2; i++) {
 7             for (int j = 0; j < 4; j++) {
 8                 for (int k = 0; k < n-1-2*i; k++) {
 9                     ans[x, y] = cur++;
10                     x += dir[j, 0];
11                     y += dir[j, 1];
12                 }
13             }
14             x++;
15             y++;
16         }
17         if (n % 2 == 1) ans[n/2, n/2] = cur;
18         return ans;
19     }
20 }

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posted @ 2013-04-21 09:32 ying_vincent 阅读(185) 评论(0) 收藏举报

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