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Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
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 link
You can find the detailed video tutorial here
It‘s a straight forward implementation question, we can simply just simulate the spiral order and print. You can choose to do it either iteratively (see reference to download the official Leetcode solution) or use recursion.
There is this awesome one line solution from this guy which is pretty insane.
def spiralOrder(self, matrix):
return matrix and list(matrix.pop(0)) + self.spiralOrder(zip(*matrix)[::-1])
https://leetcode.com/problems/spiral-matrix/discuss/20571/1-liner-in-Python-%2B-Ruby
1 public List<Integer> spiralOrder(int[][] matrix) { 2 List<Integer> res = new ArrayList<>(); 3 4 if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { 5 return res; 6 } 7 8 this.printSpiralOrder(0, 0, matrix.length, matrix[0].length, res, matrix); 9 10 return res; 11 } 12 13 // printSpiralOrder is a poor name, function starts with verb, printSpiralOrder, class is noun, function is verb 14 private void printSpiralOrder(int i, int j, int rowSize, int colSize, List<Integer> res, int[][] matrix) { 15 if (rowSize <= 0 || colSize <= 0) { 16 return; 17 } 18 19 if (rowSize == 1 && colSize == 1) { 20 res.add(matrix[i][j]); 21 return; 22 } 23 if (rowSize == 1) { 24 for (int k = j; k < j + colSize; k++) { 25 res.add(matrix[i][k]); 26 } 27 return; 28 } 29 30 if (colSize == 1) { 31 for (int k = i; k < i + rowSize; k++) { 32 res.add(matrix[k][j]); 33 } 34 return; 35 } 36 37 // do the spiral 38 for (int k = j; k < j + colSize; k++) { 39 res.add(matrix[i][k]); 40 } 41 42 for (int k = i + 1; k < i + rowSize; k++) { 43 res.add(matrix[k][j + colSize - 1]); 44 } 45 46 for (int k = j + colSize - 2; k >= i; k--) { 47 res.add(matrix[i + rowSize - 1][k]); 48 } 49 50 for (int k = i + rowSize - 2; k > i; k--) { // both the start and end need to be i, j, and also care about length 51 res.add(matrix[k][j]); 52 } 53 54 this.printSpiralOrder(i + 1, j + 1, rowSize - 2, colSize - 2, res, matrix); 55 }
Time Complexity: O(M*N) where M, N is row and col of matrix
Space Complexity: O(M*N) since we used list to store the result, where M, N is row and col of matrix
Baozi Leetcode solution 54: Sprial Matrix
标签：title return fun video vat private offic esc ber
原文地址：https://www.cnblogs.com/baozitraining/p/12015938.html
