[LeetCode] Spiral Matrix

Spiral Matrix

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

Time Complexity
O(mn)

思路

We should think about these test cases :
if matrix == null, matrix.length == 0, matrix[0] == null, matrix[0].length == 0
if matrix just has one row
1 2 3

if matrix just has one column
1
2
3
if matrix is even
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

we're going to print it in this order
1 2 3 4
8 12 16
15 14 13
9 5
6 7
11
10

if matrix is odd
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18

print it in this order
1 2 3 4 5 6
12 18
17 16 15 14 13
7

if(rowStart > rowEnd || colStart > colEnd) break; it means it just have one row or one col left, so when it finish go right and go down. It should not do go left and go down anymore.

代码

public List spiralOrder(int[][] matrix) {
    List res = new ArrayList();
    //corner case
    if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return res;
    int rows = matrix.length, cols = matrix[0].length;
    int rowStart = 0, colStart = 0, rowEnd = rows - 1, colEnd = cols - 1;
    
    while(rowStart <= rowEnd && colStart <= colEnd){
        //go right
        for(int i = colStart; i <= colEnd; i++){
            res.add(matrix[rowStart][i]);
        }
        rowStart++;
        //go down
        for(int i = rowStart; i <= rowEnd; i++){
            res.add(matrix[i][colEnd]);
        }
        colEnd--;
        
        if(rowStart > rowEnd || colStart > colEnd) break;
        
        //go left
        for(int i = colEnd; i >= colStart; i--){
            res.add(matrix[rowEnd][i]);
        }
        rowEnd--;
        
        //go up
        for(int i = rowEnd; i >= rowStart; i--){
            res.add(matrix[i][colStart]);
        }
        colStart++;
    }
    return res;
}