LeetCode: Spiral Matrix 解题报告
Spiral Matrix
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].
Solution 1:
使用递归,一次扫描一整圈,然后用x,y记录这个圈的左上角,递归完了都+1。rows, cols记录还没扫的有多少。思想比较简单,但相当容易出错。起码提交了5次才最后过。
注意:1. 扫描第一行跟最后一行要扫到底部为止,而扫描左列和右列只需要扫中间的。
1 2 3 4
5 6 7 8
9 10 11 12
例如以上例子: 你要 先扫1234, 然后是8,然后是12 11 10 9 然后是 5.
不能这样:123, 4 8, 12 11 10 , 9 5。 用后者的方法在只有一个数字 1的时候 就完全不会扫到它。
1 public List<Integer> spiralOrder1(int[][] matrix) { 2 List<Integer> ret = new ArrayList<Integer>(); 3 if (matrix == null || matrix.length == 0 4 || matrix[0].length == 0) { 5 return ret; 6 } 7 8 rec(matrix, 0, 0, matrix.length, matrix[0].length, ret); 9 10 return ret; 11 } 12 13 public static void rec(int[][] matrix, int x, int y, int rows, int cols, List<Integer> ret) { 14 if (rows <= 0 || cols <= 0) { 15 return; 16 } 17 18 // first line 19 for (int i = 0; i < cols; i++) { 20 ret.add(matrix[x][y + i]); 21 } 22 23 // right column 24 for (int i = 1; i < rows - 1; i++) { 25 ret.add(matrix[x + i][y + cols - 1]); 26 } 27 28 // down row 29 if (rows > 1) { 30 for (int i = cols - 1; i >= 0; i--) { 31 ret.add(matrix[x + rows - 1][y + i]); 32 } 33 } 34 35 // left column. GO UP. 36 if (cols > 1) { 37 for (int i = rows - 2; i > 0; i--) { 38 ret.add(matrix[x + i][y]); 39 } 40 } 41 42 rec (matrix, x + 1, y + 1, rows - 2, cols - 2, ret); 43 }
Solution 2:
http://blog.csdn.net/fightforyourdream/article/details/16876107?reload
感谢以上文章作者提供的思路,我们可以用x1,y1记录左上角,x2,y2记录右下角,这样子我们算各种边界值会方便好多。也不容易出错。
1 /* 2 Solution 2: 3 REF: http://blog.csdn.net/fightforyourdream/article/details/16876107?reload 4 此算法比较不容易算错 5 */ 6 public List<Integer> spiralOrder2(int[][] matrix) { 7 List<Integer> ret = new ArrayList<Integer>(); 8 if (matrix == null || matrix.length == 0 9 || matrix[0].length == 0) { 10 return ret; 11 } 12 13 int x1 = 0; 14 int y1 = 0; 15 16 int rows = matrix.length; 17 int cols = matrix[0].length; 18 19 while (rows >= 1 && cols >= 1) { 20 // Record the right down corner of the matrix. 21 int x2 = x1 + rows - 1; 22 int y2 = y1 + cols - 1; 23 24 // go through the WHOLE first line. 25 for (int i = y1; i <= y2; i++) { 26 ret.add(matrix[x1][i]); 27 } 28 29 // go through the right column. 30 for (int i = x1 + 1; i < x2; i++) { 31 ret.add(matrix[i][y2]); 32 } 33 34 // go through the WHOLE last row. 35 if (rows > 1) { 36 for (int i = y2; i >= y1; i--) { 37 ret.add(matrix[x2][i]); 38 } 39 } 40 41 // the left column. 42 if (cols > 1) { 43 for (int i = x2 - 1; i > x1; i--) { 44 ret.add(matrix[i][y1]); 45 } 46 } 47 48 // in one loop we deal with 2 rows and 2 cols. 49 rows -= 2; 50 cols -= 2; 51 x1++; 52 y1++; 53 } 54 55 return ret; 56 }