技術標籤：LeetcodeArray

54. Spiral Matrix

Difficulty: Medium

Related Topics: Array

Given an m x n matrix, return all elements of the matrix in spiral order.

Example 1:

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

Example 2:

Input: matrix = [[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]

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100

Solution

Key: When encountering a matrix, there are commonly some tips to think of

1. a seen matrix to note if a node has been visited: this is good so that the condition of the loop does not required to be iterated.
2. iterating through different layers, but needs to change the condition of the border of the layers when the outer loop goes for one round.

Language: Java

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int count = matrix.length * matrix[0].length - 1;
        int i =
 
 0, j = 0;
        List<Integer> result = new ArrayList<>();
        result.add(matrix[0][0]);
        int layer = 0;
        
        while (count > 0) {
            while (count > 0 && j < matrix[0].length - 1 - layer) {
                j++;
                result.add(matrix[i][j]);
                count--;
            }
            
            while (count > 0 && i < matrix.length - 1 - layer) {
                i++;
                result.add(matrix[i][j]);
                count--;
            }
            
            while (count > 0 && j > 0 + layer) {
                j--;
                result.add(matrix[i][j]);
                count--;
            }
            
            while (count > 0 && i > layer + 1) {
                i--;
                result.add(matrix[i][j]);
                count--;
            }
            layer++;
        }
        return result;
    }
}