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: