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。

 public List<Integer> spiralOrder1(int[][] matrix) {
         List<Integer> ret = new ArrayList<Integer>();
         if (matrix == null || matrix.length == 0
             || matrix[0].length == 0) {
             return ret;
         }

         rec(matrix, 0, 0, matrix.length, matrix[0].length, ret);

         return ret;
     }

     public static void rec(int[][] matrix, int x, int y, int rows, int cols, List<Integer> ret) {
         if (rows <= 0 || cols <= 0) {
             return;
         }

         // first line
         for (int i = 0; i < cols; i++) {
             ret.add(matrix[x][y + i]);
         }

         // right column
         for (int i = 1; i < rows - 1; i++) {
             ret.add(matrix[x + i][y + cols - 1]);
         }

         // down row
         if (rows > 1) {
             for (int i = cols - 1; i >= 0; i--) {
                 ret.add(matrix[x + rows - 1][y + i]);
             }
         }

         // left column. GO UP.
         if (cols > 1) {
             for (int i = rows - 2; i > 0; i--) {
                 ret.add(matrix[x + i][y]);
             }
         }

         rec (matrix, x + 1, y + 1, rows - 2, cols - 2, ret);
     }

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。

 /*
     Solution 2:
         REF: http://blog.csdn.net/fightforyourdream/article/details/16876107?reload
         此算法比较不容易算错
     */
     public List<Integer> spiralOrder2(int[][] matrix) {
         List<Integer> ret = new ArrayList<Integer>();
         if (matrix == null || matrix.length == 0
             || matrix[0].length == 0) {
             return ret;
         }

         int x1 = 0;
         int y1 = 0;

         int rows = matrix.length;
         int cols = matrix[0].length;

         while (rows >= 1 && cols >= 1) {
             // Record the right down corner of the matrix.
             int x2 = x1 + rows - 1;
             int y2 = y1 + cols - 1;

             // go through the WHOLE first line.
             for (int i = y1; i <= y2; i++) {
                 ret.add(matrix[x1][i]);
             }

             // go through the right column.
             for (int i = x1 + 1; i < x2; i++) {
                 ret.add(matrix[i][y2]);
             }

             // go through the WHOLE last row.
             if (rows > 1) {
                 for (int i = y2; i >= y1; i--) {
                     ret.add(matrix[x2][i]);
                 }
             }

             // the left column.
             if (cols > 1) {
                 for (int i = x2 - 1; i > x1; i--) {
                     ret.add(matrix[i][y1]);
                 }
             }    

             // in one loop we deal with 2 rows and 2 cols.
             rows -= 2;
             cols -= 2;
             x1++;
             y1++;
         }

         return ret;
     }

Solution 3:

http://fisherlei.blogspot.com/2013/01/leetcode-spiral-matrix.html

感谢水中的鱼大神。这是一种相当巧妙的思路，我们这次可以用Iterator来实现了，记录2个方向数组，分别表示在x方向，y方向的前进方向。1表示右或是下，-1表示左或是向上，0表示不动作。

// 1: means we are visiting the row by the right direction.
// -1: means we are visiting the row by the left direction.
int[] x = {1, 0, -1, 0};
        
// 1: means we are visiting the colum by the down direction.
// -1: means we are visiting the colum by the up direction.
int[] y = {0, 1, 0, -1};

这种方向矩阵将会很常用在各种旋转数组上。

 /*
     Solution 3:
     使用方向矩阵来求解
     */

     public List<Integer> spiralOrder(int[][] matrix) {
         List<Integer> ret = new ArrayList<Integer>();
         if (matrix == null || matrix.length == 0
             || matrix[0].length == 0) {
             return ret;
         }

         int rows = matrix.length;
         int cols = matrix[0].length;

         int visitedRows = 0;
         int visitedCols = 0;

         // indicate the direction of x    

         // 1: means we are visiting the row by the right direction.
         // -1: means we are visiting the row by the left direction.
         int[] x = {1, 0, -1, 0};

         // 1: means we are visiting the colum by the down direction.
         // -1: means we are visiting the colum by the up direction.
         int[] y = {0, 1, 0, -1};

         // 0: right, 1: down, 2: left, 3: up.
         int direct = 0;

         int startx = 0;
         int starty = 0;

         int candidateNum = 0;
         int step = 0;
         while (true) {
             if (x[direct] == 0) {
                 // visit Y axis.
                 candidateNum = rows - visitedRows;
             } else {
                 // visit X axis
                 candidateNum = cols - visitedCols;
             }

             if (candidateNum <= 0) {
                 break;
             }

             ret.add(matrix[startx][starty]);
             step++;

             if (step == candidateNum) {
                 step = 0;
                 visitedRows += x[direct] == 0 ? 0: 1;
                 visitedCols += y[direct] == 0 ? 0: 1;

                 // move forward the direction.
                 direct ++;
                 direct = direct%4;
             }

             // 根据方向来移动横坐标和纵坐标。
             startx += y[direct];
             starty += x[direct];
         }

         return ret;
     }

SOLUTION 4 (December 2nd 更新):

对solution 2改进了一下，使用top,bottom,right,left记录四个角。程序更简洁漂亮。

 public class Solution {
     public List<Integer> spiralOrder(int[][] matrix) {
         List<Integer> ret = new ArrayList<Integer>();
         if (matrix == null ||matrix.length == 0) {
             // 注意在非法的时候，应该返回空解，而不是一个NULL值
             return ret;
         }

         // Record how many rows and cols we still have.
         int rows = matrix.length;
         int cols = matrix[0].length;

         // The four coners.
         int top = 0;
         int left = 0;
         int bottom = rows - 1;
         int right = cols - 1;

         // every time we go through two rows and two cols.
         for (; rows > 0 && cols > 0; rows -= 2, cols -= 2, top++, left++, bottom--, right--) {
             // the first line.
             for (int i = left; i <= right; i++) {
                 ret.add(matrix[top][i]);
             } 

             // the right column.
             for (int i = top + 1; i < bottom; i++) {
                 ret.add(matrix[i][right]);
             }

             // the down line;
             if (rows > 1) {
                 for (int j = right; j >= left; j--) {
                     ret.add(matrix[bottom][j]);
                 }
             }

             // the left column.
             if (cols > 1) {
                 for (int i = bottom - 1; i > top; i --) {
                     ret.add(matrix[i][left]);
                 }
             }
         }

         return ret;
     }
 }

2015.1.9 redo:

可以不用rows/cols来记录行列数。只需要判断四个边界是否相撞即可。

 public List<Integer> spiralOrder3(int[][] matrix) {
         List<Integer> ret = new ArrayList<Integer>();
         if (matrix == null || matrix.length ==  || matrix[].length == ) {
             return ret;
         }

         int rows = matrix.length;
         int cols = matrix[].length;

         int left = ;
         int right = cols - ;
         int top = ;
         int bottom = rows - ;

         while (left <= right && top <= bottom) {
             // line top.
             for (int i = left; i <= right; i++) {
                 ret.add(matrix[top][i]);
             }

             // line right;
             for (int i = top + ; i <= bottom - ; i++) {
                 ret.add(matrix[i][right]);
             }

             // line bottom.
             if (top != bottom) {
                 for (int i = right; i >= left; i--) {
                     ret.add(matrix[bottom][i]);
                 }
             }

             // line left;
             if (left != right) {
                 for (int i = bottom - ; i >= top + ; i--) {
                     ret.add(matrix[i][left]);
                 }
             }

             left++;
             right--;
             top++;
             bottom--;
         }

         return ret;
     }

GitHub CODE:

spiralOrder.java