LeetCode（35）螺旋矩阵【矩阵】【中等】

news2024/12/28 3:03:50

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1.题目

给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,3,6,9,8,7,4,5]

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
输出：[1,2,3,4,8,12,11,10,9,5,6,7]

提示：

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

2.答案

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        // 遍历
        int size = matrix.length * matrix[0].length;
        List<Integer> list = new ArrayList<>(size);
        int circleIndex = 0;
        int i = 0;
        while (i < size) {
            int length = matrix[0].length - circleIndex * 2;
            for (int j = 0; j < length; j++) {
                int line = circleIndex;
                int column = circleIndex + j;
                list.add(matrix[line][column]);
                i++;
            }
            int width = matrix.length - 2 * circleIndex;
            if (width > 1) {
                for (int j = 1; j < width; j++) {
                    int line = circleIndex + j;
                    int column = circleIndex + length - 1;
                    list.add(matrix[line][column]);
                    i++;
                }
                if (length > 1) {
                    for (int j = length - 2; j >= 0; j--) {
                        int line = circleIndex + width - 1;
                        int column = circleIndex + j;
                        list.add(matrix[line][column]);
                        i++;
                    }
                    for (int j = width - 2; j > 0; j--) {
                        int line = circleIndex + j;
                        int column = circleIndex;
                        list.add(matrix[line][column]);
                        i++;
                    }
                }
            }

            circleIndex++;
        }
        return list;
    }
}