LeetCode: Spiral Matrix II 解题报告-三种方法解决旋转矩阵问题

Spiral Matrix II
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

For example,
Given n = 3,

You should return the following matrix:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

SOLUTION 1:

还是与上一题Spiral Matrix类似的算法，我们使用x1,y1作为左上角的起点，x2,y2记录右下角，这样子旋转时会简单多了。

 

 1 public int[][] generateMatrix1(int n) {

 2         int[][] ret = new int[n][n];

 3 

 4         if (n == 0) {

 5             // return a [] not a NULL.

 6             return ret;

 7         }

 8         

 9         int number = 0;

10         int rows = n;

11         

12         int x1 = 0;

13         int y1 = 0;

14         

15         while (rows > 0) {

16             int x2 = x1 + rows - 1;

17             int y2 = y1 + rows - 1;

18             

19             // the Whole first row.

20             for (int i = y1; i <= y2; i++) {

21                 number++;

22                 ret[x1][i] = number;

23             }

24             

25             // the right column except the first and last line.

26             for (int i = x1 + 1; i < x2; i++) {

27                 number++;

28                 ret[i][y2] = number;

29             }

30             

31             // This line is very important.

32             if (rows <= 1) {

33                 break;

34             }

35             

36             // the WHOLE last row.

37             for (int i = y2; i >= y1; i--) {

38                 number++;

39                 ret[x2][i] = number;

40             }

41             

42             // the left column. column keep stable

43             // x: x2-1 --> x1 + 1

44             for (int i = x2 - 1; i > x1; i--) {

45                 number++;

46                 ret[i][y1] = number;

47             }

48             

49             // remember this.

50             rows -= 2;

51             x1++;

52             y1++;

53         }

54         

55         return ret;

56     }

View Code

 

SOLUTION 2:

还是与上一题Spiral Matrix类似的算法，使用Direction 数组来定义旋转方向。其实蛮复杂的，也不好记。但是记住了应该是标准的算法。

 

 1 /*

 2         Solution 2: use direction.

 3     */

 4     public int[][] generateMatrix2(int n) {

 5         int[][] ret = new int[n][n];

 6         if (n == 0) {

 7             return ret;

 8         }

 9         

10         int[] x = {1, 0, -1, 0};

11         int[] y = {0, 1, 0, -1};

12         

13         int num = 0;

14         

15         int step = 0;

16         int candElements = 0;

17         

18         int visitedRows = 0;

19         int visitedCols = 0;

20         

21         // 0: right, 1: down, 2: left, 3: up.

22         int direct = 0;

23         

24         int startx = 0;

25         int starty = 0;

26         

27         while (true) {

28             if (x[direct] == 0) {

29                 // visit the Y axis

30                 candElements = n - visitedRows;

31             } else {

32                 // visit the X axis

33                 candElements = n - visitedCols;

34             }

35             

36             if (candElements <= 0) {

37                 break;

38             }

39             

40             // set the cell.

41             ret[startx][starty] = ++num;

42             step++;

43             

44             // change the direction.

45             if (step == candElements) {

46                 step = 0;

47                 visitedRows += x[direct] == 0 ? 0: 1;

48                 visitedCols += y[direct] == 0 ? 0: 1;

49                 

50                 // change the direction.

51                 direct = (direct + 1) % 4;

52             }

53             

54             startx += y[direct];

55             starty += x[direct];

56         }

57         

58         return ret;

59     }

View Code

 

SOLUTION 3:

无比巧妙的办法，某人的男朋友可真是牛逼啊！[leetcode] Spiral Matrix | 把一个2D matrix用螺旋方式打印

此方法的巧妙之处是使用TOP,BOOTOM, LEFT, RIGHT 四个边界条件来限制访问。其实和第一个算法类似，但是更加简洁易懂。10分钟内AC!

 

 1 /*

 2         Solution 3: 使用四条bound来限制的方法.

 3     */

 4     public int[][] generateMatrix(int n) {

 5         int[][] ret = new int[n][n];

 6         if (n == 0) {

 7             return ret;

 8         }

 9         

10         int top = 0, bottom = n - 1, left = 0, right = n - 1;

11         int num = 1;

12         while (top <= bottom) {

13             if (top == bottom) {

14                 ret[top][top] = num++;

15                 break;

16             }

17             

18             // first line.

19             for (int i = left; i < right; i++) {

20                 ret[top][i] = num++;

21             }

22             

23             // right line;

24             for (int i = top; i < bottom; i++) {

25                 ret[i][right] = num++;

26             }

27             

28             // bottom line;

29             for (int i = right; i > left; i--) {

30                 ret[bottom][i] = num++;

31             }

32             

33             // left line;

34             for (int i = bottom; i > top; i--) {

35                 ret[i][left] = num++;

36             }

37             

38             top++;

39             bottom--;

40             left++;

41             right--;

42         }

43         

44         return ret;

45     }

View Code

 

GitHub Code:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/array/GenerateMatrix1.java