Leetcode 54: Spiral Matrix Matrix spiral

54: Spiral Matrix Matrix spiral

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

Given a m X n- matrix elements ( m row, n- column), follow a helical clockwise order, returns all the elements in the matrix.

Example 1:

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

Example 2:

Input:
[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

Problem-solving ideas:

Reference Example II was observed to change the way the index: (0,0) ---> (0,3), (0,3) ---> ((2,3) ---> (2,0) - -> (1,0) ---> (1,2)

From (0,3) See, are: 1 down abscissa incremented to 2; left: Save ordinate from 1 to 0; abscissa upwardly from the minus 1 to 1; the vertical coordinate from the right by 1 to 2

If m * n matrix, from (0, m-1) starts moving down to the n-1 times the lowest, then left m-1 times, n-2 times upward, rightward m-2 times, followed by: downward n-3, left m-3, up n-4, rightward m-4. M or n are each turned from minus 1.

This is my way of thinking, the online index coordinate many of which are directly manipulated, I think it is not well understood, because more than one spiral matrix, each time to change the reference coordinates, but the two methods is not very different in nature

java：

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List nums=new ArrayList();
        if (matrix.length==0||matrix[0].length==0)return nums ;
        int row=matrix.length-1,col=matrix[0].length-1,m=0,n=0,i=-1,tmp=0;

        while (row>=0&&col>=0){
            switch (i++%4){
                case 0:
                    for (tmp=0;tmp<row;tmp++)
                        nums.add(matrix[++m][n]);row-=1;
                    break;
                case 1:
                    for (tmp=0;tmp<col;tmp++)
                        nums.add(matrix[m][--n]);col-=1;
                    break;
                case 2:
                    for (tmp=0;tmp<row;tmp++)
                        nums.add(matrix[--m][n]);row-=1;
                    break;
                case 3:
                    for (tmp=0;tmp<col;tmp++)
                        nums.add(matrix[m][++n]);col-=1;
                    break;
                default:
                    for (tmp=0;tmp<=col;tmp++)
                        nums.add(matrix[m][n++]);tmp=0;n-=1;
                    break;
            }
        }
        return nums;
    }
}

important point:

First determine whether an empty array, sequence determination condition can not be reversed. Because if the matrix.length==0determination is true, then the latter matrix[0].length==0will not be determined, i.e., returns an empty array; however, matrix[0].length==0when the front, if the input array is empty, matrix[0]will be given because no matrix index 0.

python3：

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        if len(matrix)==0 or len(matrix[0])==0:return []
        nums=[];m=0;n=0;row=len(matrix)-1;col=len(matrix[0])-1;flag=0;
        for n in range(col+1):nums.append(matrix[m][n])
        while row>=0 and col>=0:
            if flag % 4 == 0:
                for i in range(row):
                    m+=1
                    nums.append(matrix[m][n])
                row -= 1
            elif flag % 4==1:
                for i in range(col):
                    n-=1
                    nums.append(matrix[m][n])
                col -= 1
            elif flag % 4 == 2:
                for i in range(row):
                    m-=1
                    nums.append(matrix[m][n])
                row -= 1
            elif flag % 4 == 3:
                for i in range(col):
                    n+=1
                    nums.append(matrix[m][n])
                col -= 1
            flag+=1
        return nums

important point:

python does not switch ... case ... statement. cycle may for high operability, can operate directly traversal index coordinate change, it will not be repeated.