59. The two spiral matrix

Subject description:

 Given a positive integer n, to generate a 1 n2 contains all of the elements, and the element arranged spirally clockwise order square matrix.

 Example:

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

answer:

 

Generating a n × n matrix empty mat, and then fill the entire inner circumferential simulation:
the definition of the current left and right upper and lower boundaries l, r, t, b, the initial value of the num = 1, the value of the iteration termination tar = n * n;
if num <when = tar, left to right and top to bottom is always filled left to right order from the lower to the cycle, after each fill:
performing num + = 1: a need to obtain the numerical filled;
update border: after completing e.g. from left to right, the upper boundary t + = 1, corresponding to an upper boundary inward contraction.
Use num <= tar instead l <r || t <b conditions as iteration, to solve the problem when n is odd, the digital center of the matrix can not be filled in an iterative process.
Eventually return to mat.

public class L59 {
    public static int[][] generateMatrix(int n) {
        int[][] re = new int[n][n];
        int l = 0,t = 0,r = n ,b = n;
        int index = 1;
        while(index <= n *n){
            for(int i = l;i<r;i++){
                re[t][i] = index ++;
            }
            t ++;
            for(int i = t;i<b;i++){
                re[i][r-1] = index ++;
            }
            r --;
            for ( int i = r-1;i >= l;i--){
                re[b-1][i] = index ++;
            }
            b --;
            for(int i = b-1;i>=t;i--){
                re[i][l] = index ++;
            }
            l ++;
        }
        return re;
    }

    public static void main(String[] args) {
        int[][] re = generateMatrix(3);
        int s = 0;
    }
}