leetcode59 spiral matrix2

思路与54相同，由于为行=列的数组，再判断何时跳出时可简化步骤。

class Solution {
public int[][] generateMatrix(int n) {
int[][] matrix = new int[n][n];
int u=0,d=n-1,s=1;
while(d>=u){
for(int i=u;i<=d;i++) {matrix[u][i]=s++;}u++;
for(int i=u;i<=d;i++) {matrix[i][d]=s++;}//r--;
for(int i=d-1;i>=u-1;i--) {matrix[d][i]=s++;}d--;
for(int i=d;i>=u;i--) {matrix[i][u-1]=s++;}//l++;

}
return matrix;
}
}

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