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].

 

思路： 用标记法来实现循环控制，缺点会占用m*n的boolean空间，优点，简单，容易理解

用if控制循环方向，while来控制遍历该方向所有值

代码：

1. public List<Integer> spiralOrder(int[][] matrix) {
2. List<Integer> res = new ArrayList<Integer>();
3. int h = matrix.length;
4. if(h==0) return res;
5. int w = matrix[0].length;
6. if(w==0) return res;
7. boolean[][] flag = new boolean[h][w];
8. res.add(matrix[0][0]);
9. flag[0][0] = true;
10. int i=0;
11. int j=0;
12. while(true) {
13. if(j+1<w && flag[i][j+1] == false) {
14. while(j+1<w && flag[i][j+1] == false) {  //要在该方向查询完毕，否则可能有两个方向满足条件
15. flag[i][j+1] = true;
16. res.add(matrix[i][j+1]);
17. j=j+1;
18. }
19. } else if(i+1<h && flag[i+1][j] == false) {
20. while(i+1<h && flag[i+1][j] == false) {
21. flag[i+1][j] = true;
22. res.add(matrix[i+1][j]);
23. i=i+1;
24. }
25. } else if(j-1>=0 && flag[i][j-1] == false) {
26. while(j-1>=0 && flag[i][j-1] == false) {
27. flag[i][j-1] = true;
28. res.add(matrix[i][j-1]);
29. j=j-1;
30. }
31. } else if(i-1>=0 && flag[i-1][j] == false) {
32. while(i-1>=0 && flag[i-1][j] == false) {
33. flag[i-1][j] = true;
34. res.add(matrix[i-1][j]);
35. i = i-1;
36. }
37. } else {
38. break;
39. }
40. }
41. return res;
42. }

思路二： 用highrow、lowrow、lowcol、highcol四个变量控制遍历

C++代码：

1. vector<int> spiralOrder(vector<vector<int> > &matrix) {
2. int lowrow = 0;
3. int lowcol = 0;
4. vector<int> res;
5. int highrow=matrix.size();
6. if(highrow==0) return res;
7. int highcol=matrix[0].size();
8. highrow--;
9. highcol--;
10. int j=0;
11. if(highrow==0) return matrix[0];
12. while(true) {
13. if(highrow<lowrow||highcol<lowcol) break;
15. if(lowrow==highrow) {
16. for(j=lowcol;j<=highcol;j++) {
17. res.push_back(matrix[lowrow][j]);
18. }
19. break;
20. }
21. if(lowcol==highcol) {
22. for(j=lowrow;j<=highrow;j++) {
23. res.push_back(matrix[j][highcol]);
24. }
25. break;
26. }
27. for(j=lowcol;j<=highcol;j++) {
28. res.push_back(matrix[lowrow][j]);
29. }
30. for(j=lowrow+1;j<=highrow;j++) {
31. res.push_back(matrix[j][highcol]);
32. }
33. for(j=highcol-1;j>=lowcol;j--) {
34. res.push_back(matrix[highrow][j]);
35. }
36. for(j=highrow-1;j>lowrow;j--) {
37. res.push_back(matrix[j][lowcol]);
38. }
39. lowrow++;
40. lowcol++;
41. highcol--;
42. highrow--;
44. }
45. return res;
46. }