leetcode54 Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,

Given the following matrix:

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

You should return

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

.

package leetcode54;

import java.util.ArrayList;
import java.util.List;

public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> result = new ArrayList<Integer>();
int row = matrix.length;
if (row == 0) {
return result;
}
int col = matrix[0].length;
int count = 1;

int allLen = row * col;
int i = 0;
int j = 0;
boolean used[][] = new boolean[row][col];
for (int i1 = 0; i1 < row; i1++) {
for (int j1 = 0; j1 < col; j1++) {
used[i1][j1] = false;
}
}
/*for (int i1 = 0; i1 < row; i1++) {
for (int j1 = 0; j1 < col; j1++) {
System.out.println(used[i1][j1]);
}
}*/
while (count < allLen-1 ) {
while (j + 1 < col && used[i][j + 1] == false) {
result.add(matrix[i][j]);
used[i][j] = true;
j++;

count++;
}
while (i + 1 < row && used[i + 1][j] == false) {
result.add(matrix[i][j]);
used[i][j] = true;
i++;

count++;
}
while (j - 1 >= 0 && used[i][j - 1] == false) {
result.add(matrix[i][j]);
used[i][j] = true;
j--;

count++;
}
while (i - 1 >= 0 && used[i - 1][j] == false) {
result.add(matrix[i][j]);
used[i][j] = true;
i--;

count++;
}

}
for (int i1 = 0; i1 < row; i1++) {
for (int j1 = 0; j1 < col; j1++) {
if (used[i1][j1] == false) {
result.add(matrix[i1][j1]);
}
}
}
// result.add(matrix[i][j]);
return result;
}

public static void main(String[] args) {
int[][] matrix = { /* { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } */ };
int[][] matrix1 = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } };
List<Integer> result = new Solution().spiralOrder(matrix1);
for (int i = 0; i < result.size(); i++) {
System.out.print(result.get(i) + " ");
}
}

}