54 Spiral Matrix – Medium

Problem:

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].

Thoughts:

Go in a Spiral Way, one trip contains four part, go right, go down, go left and go up.

Repeat Spiral trip until there is only one column or row left.

Solutions:

``````public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> result = new ArrayList<Integer>();
if(matrix == null || matrix.length == 0) return result;
int m = matrix.length;
int n = matrix[0].length;
int x=0;
int y=0;
while(m>0 && n>0){
//if one row/column left, no circle can be formed
if(m==1){
for(int i=0; i<n; i++){
}
break;
}
else if(n==1){
for(int i=0; i<m; i++){
}
break;
}
//below, process a circle
//top - move right
for(int i=0;i<n-1;i++){
}
//right - move down
for(int i=0;i<m-1;i++){
}
//bottom - move left
for(int i=0;i<n-1;i++){
}
//left - move up
for(int i=0;i<m-1;i++){
}
x++;
y++;
m=m-2;
n=n-2;
}
return result;
}
}
``````

Alternative solution:

``````public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
if (matrix.length == 0) {
return result;
}
int left = 0, right = matrix[0].length - 1, up = 0, down = matrix.length - 1;
int i = 0, j = 0, deltaI = 0, deltaJ = 1;
while (left <= right && up <=down) {
if (j > right) {
deltaJ = 0;
deltaI = 1;
j = right;
up ++;
}
else if (i > down) {
deltaJ = -1;
deltaI = 0;
i = down;
right --;
}
else if (j < left) {
deltaJ = 0;
deltaI = -1;
j = left;
down --;
}
else if (i < up) {
deltaI = 0;
deltaJ = 1;
i = up;
left ++;
}
else {