回旋数组

O(n)级别的，一次遍历就完事，就是条件有点多

public class Solution {
    public int[][] generateMatrix(int n) {
        int num = n * n;
        int c = 0;
        int row = 1;
        int m = 1;
        int x = 0;
        int count = n;
        int[][] arr = new int[n][n];
        for(int i = 1; i <= num; i++) {

            if(c >= 0 && c < count) {
                arr[row - 1][x++] = i;
                c ++;
            }
            else if(c >= count  && c < count * 2 - 2) {
                arr[m++][n - row] = i;
                c ++;
            }else if(c >= count * 2 - 2 && c < count * 3 - 2) {
                arr[n - row][--x] = i;
                c ++;
            }else if(c >= count * 3 - 2 && c < count * 4 - 4){
                arr[--m][row - 1] = i;
                c ++;
            }
            if(c == count * 4 - 4) {
                    count = n - 2 * row;
                row++;
                x++;
                m++;
                c = 0;
            }
        }
        return arr;
    }
}