leetcode54- Spiral Matrix- medium
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]
.
每一while循环内做一次圈打印。一圈打印做4个for循环,每个for循环做一个方向。做好后更新一下某个方向的边界。
细节:中间需要检查一下是否会因为奇数窄形状导致的同条回荡。
class Solution { public List<Integer> spiralOrder(int[][] matrix) { List<Integer> result = new ArrayList<>(); if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { return result; } int m = matrix.length, n = matrix[0].length; int top = 0, bottom = m - 1, left = 0, right = n - 1; while (top <= bottom && left <= right) { for (int i = left; i <= right; i++) { result.add(matrix[top][i]); } top++; for (int i = top; i <= bottom; i++) { result.add(matrix[i][right]); } right--; // 为了避免那种窄的,单数行的最后左右回荡两次 if (top > bottom || left > right) { break; } for (int i = right; i >= left; i--) { result.add(matrix[bottom][i]); } bottom--; for (int i = bottom; i >= top; i--) { result.add(matrix[i][left]); } left++; } return result; } }