leetcode-java-54. Spiral Matrix
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/*Given a matrix of m x n elements (m rows, n columns), return all elementsof the matrix in spiral(螺旋) 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].*//*思路:(横坐标就是层,纵坐标是列)1.每螺旋一层,其实路径就是(i,i)--(i,n-1-i)--(m-1-i,n-1-i)--(m-1-i,i)--回到起点需要注意:1.循环的次数:这个需要注意(Math.min(m,n)+1)/2;2.从右到左,从下到上防止重复,则需要判断 */ public class Solution { public List<Integer> spiralOrder(int[][] matrix) { List<Integer> list = new ArrayList<Integer>(); int m = matrix.length; if(m == 0) { return list; } int n = matrix[0].length, maxLoop = (Math.min(m,n)+1)/2; for(int i = 0;i < maxLoop;i++) { // (i,i)--(i,n-1-i) for(int j = i;j <= n-1-i;j++) { list.add(matrix[i][j]); } // (i+1,n-1-i)--(m-1-i,n-1-i) for(int j = i+1;j <= m-1-i;j++) { list.add(matrix[j][n-1-i]); } // (m-1-i,n-i-2)--(m-1-i,i) if(m-1-i > i) { for(int j = n-i-2;j >= i;j--) { list.add(matrix[m-1-i][j]); } } // (m-i-2,i)--回到(i,i) if(n-1-i > i) { for(int j = m-i-2;j > i;j--) { list.add(matrix[j][i]); } } } return list; }}
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