[Leetcode 68] 59 Spiral Matrix II

Problem:

Given an integer n, generate a square matrix filled with elements from 1 to n² in spiral order.

For example,
Given n = 3,

You should return the following matrix:

[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

Analysis:

Treat it as a simulation problem. Use a "dir" variable to represent the direction according to which to fill the data. In my solution, 1 for left, 2 for right, 3 for up and 4 for down.

Then according to this direction variable, we fill in element k into the position (i, j). i and j is updated each time after filling according to the direction. But after the update, the new (i, j) position may be either a position already been visited or simple out of boundry, then it's the signal to update the direction. The direction transition is as follows:

right -> down

down -> left

left -> up

up -> right

With all these updates, we can generate the spirla matrix.

Code:

class Solution {
public:
    vector<vector<int> > generateMatrix(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > res;
        
        for (int i=0; i<n; i++) {
            vector<int> tmp(n, 0);
            res.push_back(tmp);
        }
        
        int sqr = n*n;
        
        // dir 1 -> left, 2 -> right, 3 -> up, 4 -> down
        int i=0, j=0, dir = 2;
        for (int k=1; k<=sqr; k++) {
            switch (dir) {
                case 1: { 
                    res[i][j--] = k;
                    if (j < 0 || res[i][j] != 0) {
                        j++; i--;
                        dir = 3;
                    }
                    break;
                }
                case 2: {
                    res[i][j++] = k; 
                    
                    if(j >= n || res[i][j] != 0) {
                        j--; i++;
                        dir = 4;
                    }
                    break;
                }
                case 3: {
                    res[i--][j] = k; 
                    
                    if (i < 0 || res[i][j] != 0) {
                        i++; j++;
                        dir = 2;
                    }
                    break;
                }
                case 4: {
                    res[i++][j] = k; 
                    
                    if (i >= n || res[i][j] != 0) {
                        i--; j--;
                        dir = 1;
                    }
                    break;
                }
                default: break;
            }
        }
     
        return res;
    }

};