[LeetCode] Spiral Matrix II
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3
,
[ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ]
Solution:
class Solution { public: int **used; vector<vector<int> > generateMatrix(int n) { vector< vector<int > > ans; if(n == 0) return ans; used = new int*[n]; for(int i = 0;i < n;i++) { used[i] = new int[n]; memset(used[i], 0, n * sizeof(int)); } int x = 0, y = -1, count = 1, dir = 1; //1 for right, 2 for down, 3 for left, 4 for up bool flag = true; while(flag) { switch(dir) { case 1: //current direction is right, try continue if(y + 1 < n && used[x][y + 1] == 0) used[x][(y++) + 1] = count++; else if(x + 1 < n && used[x + 1][y] == 0) { used[(x++) + 1][y] = count++; dir = 2; } else flag = false; break; case 2: //current direction is right, try continue if(x + 1 < n && used[x + 1][y] == 0) used[(x++) + 1][y] = count++; else if(y - 1 >= 0 && used[x][y - 1] == 0) { used[x][(y--) - 1] = count++; dir = 3; } else flag = false; break; case 3: //current direction is right, try continue if(y - 1 >= 0 && used[x][y - 1] == 0) used[x][(y--) - 1] = count++; else if(x - 1 >= 0 && used[x - 1][y] == 0) { used[(x--) - 1][y] = count++; dir = 4; } else flag = false; break; case 4: //current direction is right, try continue if(x - 1 >= 0 && used[x - 1][y] == 0) used[(x--) - 1][y] = count++; else if(y + 1 < n && used[x][y + 1] == 0) { used[x][(y++) + 1] = count++; dir = 1; } else flag = false; break; default: break; } } for(int i = 0;i < n;i++) { vector<int > tmp; for(int j = 0;j < n;j++) tmp.push_back(used[i][j]); ans.push_back(tmp); } return ans; } };