标签:style blog http color io ar for sp div
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example, Given the following matrix: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] You should return [1,2,3,6,9,8,7,4,5].
难度:87,这道题跟 Rotate Image 很相似,都是需要把矩阵分割成层来处理,每一层都是按:1. 正上方;2. 正右方;3. 正下方;4. 正左方这种顺序添加元素到结果集合。实现中要注意两个细节,一个是因为题目中没有说明矩阵是不是方阵,因此要先判断一下行数和列数来确定螺旋的层数,mina(行, 列) /2才是螺旋数。二是有可能存在这种情况比如:3*4矩阵,螺旋只有一层,但中间还有一行元素没有没加入。所以最后要将剩余的走完。
1 public class Solution { 2 public List<Integer> spiralOrder(int[][] matrix) { 3 ArrayList<Integer> res = new ArrayList<Integer>(); 4 if (matrix==null || matrix.length==0 || matrix[0].length==0) return res; 5 int Min = Math.min(matrix.length, matrix[0].length); 6 int levelMax = Min / 2; 7 for (int level=0; level<levelMax; level++) { 8 for (int i=level; i<matrix[0].length-1-level; i++) { 9 res.add(matrix[level][i]); 10 } 11 for (int i=level; i<matrix.length-1-level; i++) { 12 res.add(matrix[i][matrix[0].length-1-level]); 13 } 14 for (int i=matrix[0].length-1-level; i>level; i--) { 15 res.add(matrix[matrix.length-1-level][i]); 16 } 17 for (int i=matrix.length-1-level; i>level; i--) { 18 res.add(matrix[i][level]); 19 } 20 } 21 if (Min % 2 == 1) { 22 if (matrix.length < matrix[0].length) { 23 for (int i=levelMax; i<matrix[0].length-levelMax; i++) { 24 res.add(matrix[levelMax][i]); 25 } 26 } 27 else { 28 for (int i=levelMax; i<matrix.length-levelMax; i++) { 29 res.add(matrix[i][levelMax]); 30 } 31 } 32 } 33 return res; 34 } 35 }
标签:style blog http color io ar for sp div
原文地址:http://www.cnblogs.com/EdwardLiu/p/4018225.html