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]
- package TOP11_20;
-
- import java.util.ArrayList;
- import java.util.List;
-
- //螺旋矩阵
- /*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]
- */
- public class Top19 {
- // 转圈遍历
- public static List
spiralOrder(int[][] matrix) { - List
result = new ArrayList<>(); - if (matrix.length == 0) {
- return result;
- }
- int length = matrix.length;
- int height = matrix[0].length;
- int totalNums = length * height;
- int left = 0;
- int right = height - 1;
- int top = 0;
- int bottom = length - 1;
- while (totalNums > 0) {
- for (int i = left; i <= right && totalNums > 0; i++) {
- result.add(matrix[top][i]);
- totalNums--;
- }
- top++;
- for (int i = top; i <= bottom && totalNums > 0; i++) {
- result.add(matrix[i][right]);
- totalNums--;
- }
- right--;
- for (int i = right; i >= left && totalNums > 0; i--) {
- result.add(matrix[bottom][i]);
- totalNums--;
- }
- bottom--;
- for (int i = bottom; i >= top && totalNums > 0; i--) {
- result.add(matrix[i][left]);
- totalNums--;
- }
- left++;
- }
- return result;
- }
-
- public static void main(String[] args) {
- int[][] matrix = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
- System.out.println(spiralOrder(matrix));
- }
- }