54螺旋矩阵

给定一个包含 m x n 个元素的矩阵(m 行, n 列),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:
[
 [ 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:
[
  [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]

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/spiral-matrix 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

Solution 1

顺时针遍历
注意当行列数短的一方为奇数的时候,只用走半圈

import java.util.*;

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        if (matrix.length == 0)
            return new LinkedList<Integer>();
        int rowLen = matrix.length;
        int colLen = matrix[0].length;
        int rank = rowLen < colLen ? (rowLen + 1) / 2 : (colLen + 1) / 2;
        boolean lastHalfCircle = rowLen < colLen ? rowLen % 2 == 1 : colLen % 2 == 1;
        List<Integer> result = new LinkedList<Integer>();

        for (int n = 0; n < rank; n++) {
            for (int i = n; i < colLen - n; i++)
                result.add(matrix[n][i]);
            for (int i = n + 1; i < rowLen - n; i++)
                result.add(matrix[i][colLen - n - 1]);
            if (n == rank - 1 && lastHalfCircle)
                break;
            for (int i = colLen - n - 2; i >= n; i--)
                result.add(matrix[rowLen - n - 1][i]);
            for (int i = rowLen - n - 2; i > n; i--)
                result.add(matrix[i][n]);
        }
        return result;
    }
}