54. Spiral Matrix

难度: Medium

题目描述

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

 

Example 1:

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

解题思路

代码实现

解决方案

java

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int row = matrix.length;
        int col = matrix[0].length;
        int[][] record = new int[row][col];
        List<Integer> res = new ArrayList<>();
        int mode = 0;
        int rowIndex = 0;
        int colIndex = 0;
        int size = row * col;
        while (res.size() < size) {

            if (mode % 4 == 0) {
                for (int i = colIndex; i < col && record[rowIndex][i] != 1; i++) {
                    res.add(matrix[rowIndex][i]);
                    record[rowIndex][i] = 1;
                    colIndex = i;
                }
                rowIndex++;

            } else if (mode % 4 == 1) {
                for (int i = rowIndex; i < row && record[i][colIndex] != 1; i++) {
                    res.add(matrix[i][colIndex]);
                    record[i][colIndex] = 1;
                    rowIndex=i;
                }
                colIndex--;

            } else if (mode % 4 == 2) {
                for (int i = colIndex; i > -1 && record[rowIndex][i] != 1; i--) {
                    res.add(matrix[rowIndex][i]);
                    record[rowIndex][i] = 1;
                    colIndex = i;
                }
                rowIndex--;

            } else {
                for (int i = rowIndex; i > -1 && record[i][colIndex] != 1; i--) {
                    res.add(matrix[i][colIndex]);
                    record[i][colIndex] = 1;
                    rowIndex=i;
                }
                colIndex++;
            }
            mode++;

        }

        return res;

    }
}