Leetcode54. 螺旋矩阵
问题描述:
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
上代码,拿去即可运行:
package com.onlyqi.daydayupgo01.test;import java.util.ArrayList;
import java.util.List;public class Test28 {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));}public static List<Integer> spiralOrder(int[][] matrix) {List<Integer> integerList = new ArrayList<>();int rows = matrix.length, columns = matrix[0].length;int left = 0, right = columns - 1, bottom = rows - 1, top = 0;// 列行行列---》也即:横-竖-反向横-反向竖while (left <= right && top <= bottom) {for (int column = left; column <= right; column++) {integerList.add(matrix[top][column]);}for (int row = top + 1; row <= bottom; row++) {integerList.add(matrix[row][right]);}if (left < right && top < bottom) {for (int column = right - 1; column > left; column--) {integerList.add(matrix[bottom][column]);}for (int row = bottom; row > top; row--) {integerList.add(matrix[row][left]);}}left++;right--;top++;bottom--;}return integerList;}}
慢慢来才是最快的方法--天涯明月 共勉
我要刷300道算法题,第135道 。 希望自己可以坚持下去 。