Leetcode54螺旋矩阵
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;
}
}