Leetcode-59-Spiral Matrix II

Spiral Matrix II

leetcode: https://leetcode.com/problems/spiral-matrix-ii/

Description:

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

Idea:

solution1:

Step1: set two index for the round(one for row and the other for column), and index of the number of times to loop

Step2: the number of the rounds we need take is n/2. if the number is odd, we need to assign the value to the array central, if the number if even, when all rounds finished, all the elements will be assigned.

Step3: set two index of loop and set the same interval rule of the loop for each side of the square, for example, in each side only include from the first element to penultimate element, and left the last element for next side(as the first element)

Code:

class Solution {
    public int[][] generateMatrix(int n) {
        //initialize the result array
        int[][] result = new int[n][n];
        //set the bound index for round
        int bound_i = 0, bound_j =0;
        //set the index to reocrd the number of times it has looped
        int loop = 0;
        //to give each element value with self-increase it 
        int count = 1;
        // set two index for the loop
        int i,j;
        //n/2 means the number of times it needed
        while(loop < n/2){
            //make two loop index point to the correct bound of the current round
            i=bound_i;
            j=bound_j;
            //upper side, and just move the column index to assign the value to all element in this row
            //each side don't deal with the last element and left them as the first element for the next side
            for(j=bound_j; j<n-1-loop; j++){
                result[i][j] = count++;
            }
            //right side, just move row index, and because the last loop has move the column index to the right index.
            for(i=bound_i; i<n-1-loop; i++){
                result[i][j] = count++;
            }
            //bottom side, move it back to the start bound of this round
            for(;j>bound_j; j--){
                result[i][j] = count++;
            }
            //left side
            for(;i>bound_i; i--){
                result[i][j] = count++;
            }
            //narrowing the bound to next round
            bound_i++;
            bound_j++;
            loop++;
        }
        if (n%2 == 1){
            result[n/2][n/2] = count++;
        }
        return result;
    }
}