LeetCode: Spiral Matrix 解题报告

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转载网址：http://www.tuicool.com/articles/bA3yAf

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 line19         for (int i = 0; i < cols; i++) {20             ret.add(matrix[x][y + i]);21         }22         23         // right column24         for (int i = 1; i < rows - 1; i++) {25             ret.add(matrix[x + i][y + cols - 1]);26         }27         28         // down row29         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     }

View Code

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     }

View Code

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。

 1 /* 2     Solution 3: 3     使用方向矩阵来求解 4     */ 5      6     public List<Integer> spiralOrder(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 rows = matrix.length;14         int cols = matrix[0].length;15         16         int visitedRows = 0;17         int visitedCols = 0;18 19         // indicate the direction of x    20         21         // 1: means we are visiting the row by the right direction.22         // -1: means we are visiting the row by the left direction.23         int[] x = {1, 0, -1, 0};24         25         // 1: means we are visiting the colum by the down direction.26         // -1: means we are visiting the colum by the up direction.27         int[] y = {0, 1, 0, -1};28         29         // 0: right, 1: down, 2: left, 3: up.30         int direct = 0;31         32         int startx = 0;33         int starty = 0;34         35         int candidateNum = 0;36         int step = 0;37         while (true) {38             if (x[direct] == 0) {39                 // visit Y axis.40                 candidateNum = rows - visitedRows;41             } else {42                 // visit X axis43                 candidateNum = cols - visitedCols;44             }45             46             if (candidateNum <= 0) {47                 break;48             }49             50             ret.add(matrix[startx][starty]);51             step++;52             53             if (step == candidateNum) {54                 step = 0;55                 visitedRows += x[direct] == 0 ? 0: 1;56                 visitedCols += y[direct] == 0 ? 0: 1;57                 58                 // move forward the direction.59                 direct ++;60                 direct = direct%4;61             }62             63             // 根据方向来移动横坐标和纵坐标。64             startx += y[direct];65             starty += x[direct];66         }67         68         return ret;69     }

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