示例:
I
0 0 0 0 1 0
0 0 1 0 0 1
II->0 0 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 1 <--IV
0 0 1 0 0 0
IV
对于上面的示例,它是一个6×6的二进制矩阵。在这种情况下,返回值将为Cell 1:(2,1)和Cell 2:(4,4)。所得的子矩阵可以是正方形或矩形。返回值也可以是所有0的最大子矩阵的大小,在此示例中为3×4。
#1 楼
这是一个基于@j_random_hacker在评论中建议的“直方图中的最大矩形”问题的解决方案:[算法]的工作原理是从上到下迭代
行,对于解决此问题的每一行
,其中“直方图”中的“条”由
从当前行开始的所有零连续向上的零迹
组成(
列的高度为0(如果当前行为1英寸)。
输入矩阵
mat
可以是任意可迭代的,例如文件或网络流。一次只需要一行。 #!/usr/bin/env python
from collections import namedtuple
from operator import mul
Info = namedtuple('Info', 'start height')
def max_size(mat, value=0):
"""Find height, width of the largest rectangle containing all `value`'s."""
it = iter(mat)
hist = [(el==value) for el in next(it, [])]
max_size = max_rectangle_size(hist)
for row in it:
hist = [(1+h) if el == value else 0 for h, el in zip(hist, row)]
max_size = max(max_size, max_rectangle_size(hist), key=area)
return max_size
def max_rectangle_size(histogram):
"""Find height, width of the largest rectangle that fits entirely under
the histogram.
"""
stack = []
top = lambda: stack[-1]
max_size = (0, 0) # height, width of the largest rectangle
pos = 0 # current position in the histogram
for pos, height in enumerate(histogram):
start = pos # position where rectangle starts
while True:
if not stack or height > top().height:
stack.append(Info(start, height)) # push
elif stack and height < top().height:
max_size = max(max_size, (top().height, (pos - top().start)),
key=area)
start, _ = stack.pop()
continue
break # height == top().height goes here
pos += 1
for start, height in stack:
max_size = max(max_size, (height, (pos - start)), key=area)
return max_size
def area(size):
return reduce(mul, size)
解决方案是
O(N)
,其中N
是矩阵中元素的数量。它需要O(ncols)
额外的内存,其中ncols
是矩阵中的列数。带有测试的最新版本位于https://gist.github.com/776423
评论
尝试不错,但是max_size([[0,0,0,0,0,1,1,1,[0,0,0,0,0,0,0,0,0,0,0,1, 1,1,1],[0,0,1,1,1,1,1]] + [[1,0,1,1,1,1,1,1]] * 3),返回(2,4 ),则在左上角有一个3x3的正方形。
– j_random_hacker
2011-1-14的2:02
基本的问题是,像在这里所做的那样,仅跟踪相邻点的(几个)最大面积的矩形并不总是足够的。我知道是正确的唯一O(N)算法是通过从上到下迭代遍历每一行来解决此问题的:stackoverflow.com/questions/4311694/…,其中“直方图”中的“条”由从当前行开始的所有零连续向上的轨迹组成(如果列在当前行中为1,则其高度为0)。
– j_random_hacker
2011-1-14的2:08
@j_random_hacker:我已经更新了答案,以使用基于“直方图”的算法。
– jfs
2011年1月14日12:45
看起来不错,但是,我实际上是在寻找最大的矩形(例如,返回坐标)。该算法将可靠地返回该区域,但是一旦我知道了,一个人怎么会发现3列x 2行矩形的位置,其左上角为[3,5]?
– JBWhitmore
2014年1月26日,3:25
从哪里获得绑定列信息? (矩形的左列或右列?)。我们可以从max_rectangle_size获取宽度和高度,并从for行的最下面一行获取:迭代,但是我找不到边界列信息。
–manatttta
15年6月20日在8:53
#2 楼
请查看“直方图”下的“最大化矩形区域”,然后继续阅读下面的解决方案。Traverse the matrix once and store the following;
For x=1 to N and y=1 to N
F[x][y] = 1 + F[x][y-1] if A[x][y] is 0 , else 0
Then for each row for x=N to 1
We have F[x] -> array with heights of the histograms with base at x.
Use O(N) algorithm to find the largest area of rectangle in this histogram = H[x]
From all areas computed, report the largest.
时间复杂度为O(N * N)= O(N²)(对于NxN二进制矩阵)
示例:
Initial array F[x][y] array
0 0 0 0 1 0 1 1 1 1 0 1
0 0 1 0 0 1 2 2 0 2 1 0
0 0 0 0 0 0 3 3 1 3 2 1
1 0 0 0 0 0 0 4 2 4 3 2
0 0 0 0 0 1 1 5 3 5 4 0
0 0 1 0 0 0 2 6 0 6 5 1
For x = N to 1
H[6] = 2 6 0 6 5 1 -> 10 (5*2)
H[5] = 1 5 3 5 4 0 -> 12 (3*4)
H[4] = 0 4 2 4 3 2 -> 10 (2*5)
H[3] = 3 3 1 3 2 1 -> 6 (3*2)
H[2] = 2 2 0 2 1 0 -> 4 (2*2)
H[1] = 1 1 1 1 0 1 -> 4 (1*4)
The largest area is thus H[5] = 12
评论
很好的例子说明
–彼得
13年1月22日在15:00
您确定这是O(N * N)吗?整个矩阵有两遍,但我的印象是这是O(N)。
–克里斯·梅斯(Chris Maes)
2014年3月21日在10:26
很好的解释.. :)我希望,您也已经解释了“在直方图下最大化矩形区域” ..:D
–tumbudu
2015年10月2日,15:26
更清楚地说。解为O(N * N),其中N是行/列中的项目数,因为问题指出输入的大小为NxN。如果N是输入项的总数,则为O(N)
–user2469515
16 Apr 17'2:19
#3 楼
这是一个Python3解决方案,该解决方案除了返回最大矩形区域之外,还返回位置:输出:
area 12
Cell 1:(2, 1) and Cell 2:(4, 4)
评论
很棒!我从中制作出了一个Fortran版本,并将其编译为可在Python中使用,因为这样遍历Python中的大型数组非常缓慢。
–詹森
19-6-25在2:24
#4 楼
这是将JF Sebastians方法转换为C#:private Vector2 MaxRectSize(int[] histogram) {
Vector2 maxSize = Vector2.zero;
int maxArea = 0;
Stack<Vector2> stack = new Stack<Vector2>();
int x = 0;
for (x = 0; x < histogram.Length; x++) {
int start = x;
int height = histogram[x];
while (true) {
if (stack.Count == 0 || height > stack.Peek().y) {
stack.Push(new Vector2(start, height));
} else if(height < stack.Peek().y) {
int tempArea = (int)(stack.Peek().y * (x - stack.Peek().x));
if(tempArea > maxArea) {
maxSize = new Vector2(stack.Peek().y, (x - stack.Peek().x));
maxArea = tempArea;
}
Vector2 popped = stack.Pop();
start = (int)popped.x;
continue;
}
break;
}
}
foreach (Vector2 data in stack) {
int tempArea = (int)(data.y * (x - data.x));
if(tempArea > maxArea) {
maxSize = new Vector2(data.y, (x - data.x));
maxArea = tempArea;
}
}
return maxSize;
}
public Vector2 GetMaximumFreeSpace() {
// STEP 1:
// build a seed histogram using the first row of grid points
// example: [true, true, false, true] = [1,1,0,1]
int[] hist = new int[gridSizeY];
for (int y = 0; y < gridSizeY; y++) {
if(!invalidPoints[0, y]) {
hist[y] = 1;
}
}
// STEP 2:
// get a starting max area from the seed histogram we created above.
// using the example from above, this value would be [1, 1], as the only valid area is a single point.
// another example for [0,0,0,1,0,0] would be [1, 3], because the largest area of contiguous free space is 3.
// Note that at this step, the heigh fo the found rectangle will always be 1 because we are operating on
// a single row of data.
Vector2 maxSize = MaxRectSize(hist);
int maxArea = (int)(maxSize.x * maxSize.y);
// STEP 3:
// build histograms for each additional row, re-testing for new possible max rectangluar areas
for (int x = 1; x < gridSizeX; x++) {
// build a new histogram for this row. the values of this row are
// 0 if the current grid point is occupied; otherwise, it is 1 + the value
// of the previously found historgram value for the previous position.
// What this does is effectly keep track of the height of continous avilable spaces.
// EXAMPLE:
// Given the following grid data (where 1 means occupied, and 0 means free; for clairty):
// INPUT: OUTPUT:
// 1.) [0,0,1,0] = [1,1,0,1]
// 2.) [0,0,1,0] = [2,2,0,2]
// 3.) [1,1,0,1] = [0,0,1,0]
//
// As such, you'll notice position 1,0 (row 1, column 0) is 2, because this is the height of contiguous
// free space.
for (int y = 0; y < gridSizeY; y++) {
if(!invalidPoints[x, y]) {
hist[y] = 1 + hist[y];
} else {
hist[y] = 0;
}
}
// find the maximum size of the current histogram. If it happens to be larger
// that the currently recorded max size, then it is the new max size.
Vector2 maxSizeTemp = MaxRectSize(hist);
int tempArea = (int)(maxSizeTemp.x * maxSizeTemp.y);
if (tempArea > maxArea) {
maxSize = maxSizeTemp;
maxArea = tempArea;
}
}
// at this point, we know the max size
return maxSize;
}
关于此的几点注意事项:
此版本旨在用于Unity API。通过使用KeyValuePair替换Vector2的实例,可以轻松地使其更通用。 Vector2仅用于方便的方式来存储两个值。
invalidPoints []是布尔数组,其中true表示网格点为“使用中”,false表示不是。
#5 楼
具有空间复杂度O(列)的解决方案[也可以修改为O(行)]和时间复杂度O(行*列)public int maximalRectangle(char[][] matrix) {
int m = matrix.length;
if (m == 0)
return 0;
int n = matrix[0].length;
int maxArea = 0;
int[] aux = new int[n];
for (int i = 0; i < n; i++) {
aux[i] = 0;
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
aux[j] = matrix[i][j] - '0' + aux[j];
maxArea = Math.max(maxArea, maxAreaHist(aux));
}
}
return maxArea;
}
public int maxAreaHist(int[] heights) {
int n = heights.length;
Stack<Integer> stack = new Stack<Integer>();
stack.push(0);
int maxRect = heights[0];
int top = 0;
int leftSideArea = 0;
int rightSideArea = heights[0];
for (int i = 1; i < n; i++) {
if (stack.isEmpty() || heights[i] >= heights[stack.peek()]) {
stack.push(i);
} else {
while (!stack.isEmpty() && heights[stack.peek()] > heights[i]) {
top = stack.pop();
rightSideArea = heights[top] * (i - top);
leftSideArea = 0;
if (!stack.isEmpty()) {
leftSideArea = heights[top] * (top - stack.peek() - 1);
} else {
leftSideArea = heights[top] * top;
}
maxRect = Math.max(maxRect, leftSideArea + rightSideArea);
}
stack.push(i);
}
}
while (!stack.isEmpty()) {
top = stack.pop();
rightSideArea = heights[top] * (n - top);
leftSideArea = 0;
if (!stack.isEmpty()) {
leftSideArea = heights[top] * (top - stack.peek() - 1);
} else {
leftSideArea = heights[top] * top;
}
maxRect = Math.max(maxRect, leftSideArea + rightSideArea);
}
return maxRect;
}
但是我得到了Time Limite当我在LeetCode上尝试时超出了预期。有没有更简单的解决方案?
评论
简单易懂..谢谢!
–斯瓦迪卡
16年7月9日在4:02
#6 楼
我提出了O(nxn)方法。首先,您可以列出所有最大的空矩形。空表示仅覆盖0。最大的空白矩形不能在不覆盖(至少)一个1的方向上扩展。可在www.www.analog.com/上找到介绍O(nxn)算法以创建此类列表的论文。 .ulg.ac.be / telecom / rectangles以及源代码(未优化)。无需存储列表,每次在算法找到一个矩形时调用一次回调函数就足够了,并且仅存储最大的一个(如果需要,可以选择另一个条件)。
请注意,存在一个证明(请参阅论文),其中最大的空白矩形的数量由图像的像素数量(在这种情况下为nxn)限制。
因此,选择最佳矩形可以在O(nxn)中完成,整个方法也是O(nxn)。
在实践中,这种方法非常快,用于实时视频流。分析。
#7 楼
这是jfs解决方案的一个版本,还提供了最大矩形的位置:from collections import namedtuple
from operator import mul
Info = namedtuple('Info', 'start height')
def max_rect(mat, value=0):
"""returns (height, width, left_column, bottom_row) of the largest rectangle
containing all `value`'s.
Example:
[[0, 0, 0, 0, 0, 0, 0, 0, 3, 2],
[0, 4, 0, 2, 4, 0, 0, 1, 0, 0],
[1, 0, 1, 0, 0, 0, 3, 0, 0, 4],
[0, 0, 0, 0, 4, 2, 0, 0, 0, 0],
[0, 0, 0, 2, 0, 0, 0, 0, 0, 0],
[4, 3, 0, 0, 1, 2, 0, 0, 0, 0],
[3, 0, 0, 0, 2, 0, 0, 0, 0, 4],
[0, 0, 0, 1, 0, 3, 2, 4, 3, 2],
[0, 3, 0, 0, 0, 2, 0, 1, 0, 0]]
gives: (3, 4, 6, 5)
"""
it = iter(mat)
hist = [(el==value) for el in next(it, [])]
max_rect = max_rectangle_size(hist) + (0,)
for irow,row in enumerate(it):
hist = [(1+h) if el == value else 0 for h, el in zip(hist, row)]
max_rect = max(max_rect, max_rectangle_size(hist) + (irow+1,), key=area)
# irow+1, because we already used one row for initializing max_rect
return max_rect
def max_rectangle_size(histogram):
stack = []
top = lambda: stack[-1]
max_size = (0, 0, 0) # height, width and start position of the largest rectangle
pos = 0 # current position in the histogram
for pos, height in enumerate(histogram):
start = pos # position where rectangle starts
while True:
if not stack or height > top().height:
stack.append(Info(start, height)) # push
elif stack and height < top().height:
max_size = max(max_size, (top().height, (pos - top().start), top().start), key=area)
start, _ = stack.pop()
continue
break # height == top().height goes here
pos += 1
for start, height in stack:
max_size = max(max_size, (height, (pos - start), start), key=area)
return max_size
def area(size):
return size[0] * size[1]
#8 楼
为了完整起见,这是输出矩形坐标的C#版本。它基于dmarra的答案,但没有任何其他依赖性。
只有布尔函数GetPixel(int x,int y),当在坐标x,y处设置像素时返回true。 />
public struct INTRECT
{
public int Left, Right, Top, Bottom;
public INTRECT(int aLeft, int aTop, int aRight, int aBottom)
{
Left = aLeft;
Top = aTop;
Right = aRight;
Bottom = aBottom;
}
public int Width { get { return (Right - Left + 1); } }
public int Height { get { return (Bottom - Top + 1); } }
public bool IsEmpty { get { return Left == 0 && Right == 0 && Top == 0 && Bottom == 0; } }
public static bool operator ==(INTRECT lhs, INTRECT rhs)
{
return lhs.Left == rhs.Left && lhs.Top == rhs.Top && lhs.Right == rhs.Right && lhs.Bottom == rhs.Bottom;
}
public static bool operator !=(INTRECT lhs, INTRECT rhs)
{
return !(lhs == rhs);
}
public override bool Equals(Object obj)
{
return obj is INTRECT && this == (INTRECT)obj;
}
public bool Equals(INTRECT obj)
{
return this == obj;
}
public override int GetHashCode()
{
return Left.GetHashCode() ^ Right.GetHashCode() ^ Top.GetHashCode() ^ Bottom.GetHashCode();
}
}
public INTRECT GetMaximumFreeRectangle()
{
int XEnd = 0;
int YStart = 0;
int MaxRectTop = 0;
INTRECT MaxRect = new INTRECT();
// STEP 1:
// build a seed histogram using the first row of grid points
// example: [true, true, false, true] = [1,1,0,1]
int[] hist = new int[Height];
for (int y = 0; y < Height; y++)
{
if (!GetPixel(0, y))
{
hist[y] = 1;
}
}
// STEP 2:
// get a starting max area from the seed histogram we created above.
// using the example from above, this value would be [1, 1], as the only valid area is a single point.
// another example for [0,0,0,1,0,0] would be [1, 3], because the largest area of contiguous free space is 3.
// Note that at this step, the heigh fo the found rectangle will always be 1 because we are operating on
// a single row of data.
Tuple<int, int> maxSize = MaxRectSize(hist, out YStart);
int maxArea = (int)(maxSize.Item1 * maxSize.Item2);
MaxRectTop = YStart;
// STEP 3:
// build histograms for each additional row, re-testing for new possible max rectangluar areas
for (int x = 1; x < Width; x++)
{
// build a new histogram for this row. the values of this row are
// 0 if the current grid point is occupied; otherwise, it is 1 + the value
// of the previously found historgram value for the previous position.
// What this does is effectly keep track of the height of continous avilable spaces.
// EXAMPLE:
// Given the following grid data (where 1 means occupied, and 0 means free; for clairty):
// INPUT: OUTPUT:
// 1.) [0,0,1,0] = [1,1,0,1]
// 2.) [0,0,1,0] = [2,2,0,2]
// 3.) [1,1,0,1] = [0,0,1,0]
//
// As such, you'll notice position 1,0 (row 1, column 0) is 2, because this is the height of contiguous
// free space.
for (int y = 0; y < Height; y++)
{
if (!GetPixel(x, y))
{
hist[y]++;
}
else
{
hist[y] = 0;
}
}
// find the maximum size of the current histogram. If it happens to be larger
// that the currently recorded max size, then it is the new max size.
Tuple<int, int> maxSizeTemp = MaxRectSize(hist, out YStart);
int tempArea = (int)(maxSizeTemp.Item1 * maxSizeTemp.Item2);
if (tempArea > maxArea)
{
maxSize = maxSizeTemp;
maxArea = tempArea;
MaxRectTop = YStart;
XEnd = x;
}
}
MaxRect.Left = XEnd - maxSize.Item1 + 1;
MaxRect.Top = MaxRectTop;
MaxRect.Right = XEnd;
MaxRect.Bottom = MaxRectTop + maxSize.Item2 - 1;
// at this point, we know the max size
return MaxRect;
}
private Tuple<int, int> MaxRectSize(int[] histogram, out int YStart)
{
Tuple<int, int> maxSize = new Tuple<int, int>(0, 0);
int maxArea = 0;
Stack<Tuple<int, int>> stack = new Stack<Tuple<int, int>>();
int x = 0;
YStart = 0;
for (x = 0; x < histogram.Length; x++)
{
int start = x;
int height = histogram[x];
while (true)
{
if (stack.Count == 0 || height > stack.Peek().Item2)
{
stack.Push(new Tuple<int, int>(start, height));
}
else if (height < stack.Peek().Item2)
{
int tempArea = (int)(stack.Peek().Item2 * (x - stack.Peek().Item1));
if (tempArea > maxArea)
{
YStart = stack.Peek().Item1;
maxSize = new Tuple<int, int>(stack.Peek().Item2, (x - stack.Peek().Item1));
maxArea = tempArea;
}
Tuple<int, int> popped = stack.Pop();
start = (int)popped.Item1;
continue;
}
break;
}
}
foreach (Tuple<int, int> data in stack)
{
int tempArea = (int)(data.Item2 * (x - data.Item1));
if (tempArea > maxArea)
{
YStart = data.Item1;
maxSize = new Tuple<int, int>(data.Item2, (x - data.Item1));
maxArea = tempArea;
}
}
return maxSize;
}
评论
请考虑将接受的答案更改为J.F. Sebastian的答案,该答案现在是正确的,并且具有最佳的复杂性。请检查非常相似(我会说重复)的问题:stackoverflow.com/questions/7770945/…,stackoverflow.com/a/7353193/684229。解是O(n)。
我正在尝试对任何方向的矩形执行相同的操作。看到问题:stackoverflow.com/questions/22604043/…
@TMS实际上是相反的方式。这些问题是这一问题的重复。