LeetCode刷题笔录Spiral Matrix
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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]
.
这题自己想的太复杂了。看了些别人的答案,发现最好的办法就是建立一个m*n的boolean矩阵,表示这个index是不是已经被访问过了,一开始全都是false。我们按照right, down, left, up的顺序来访问整个矩阵。在访问下一个元素之前先查看boolean矩阵看看这个元素是不是已经被访问过了。如果是那么就换方向。
public class Solution { public List<Integer> spiralOrder(int[][] matrix) { List<Integer> res = new ArrayList<Integer>(); if(matrix.length == 0) return res; int m = matrix.length; int n = matrix[0].length; boolean[][] mask = new boolean[m][n]; int i = 0, j = 0, k = 0; //add the first element res.add(matrix[0][0]); //this step is important, don't forget!!!! mask[0][0] = true; while(k < m * n - 1){ //go to the right while((j + 1 < n) && mask[i][j + 1] == false){ j++; k++; res.add(matrix[i][j]); mask[i][j] = true; } //go down while((i + 1 < m) && mask[i + 1][j] == false){ i++; k++; res.add(matrix[i][j]); mask[i][j] = true; } //go to the left while((j > 0) && mask[i][j - 1] == false){ j--; k++; res.add(matrix[i][j]); mask[i][j] = true; } //go up while((i > 0) && mask[i - 1][j] == false){ i--; k++; res.add(matrix[i][j]); mask[i][j] = true; } } return res; }}
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