输入: 一个N*N的二维数组矩阵,矩阵中包含正数和负数元素。
输出: 任意大小的子矩阵,其所有子矩阵中元素之和最大。
要求: 算法复杂度为O(N^3)
历史: 在Algorithmist Larry和Kadane算法的修改帮助下,我已经解决了部分问题,即仅确定总和 - 下面是Java代码。
感谢Ernesto解决了该问题的其余部分,即确定矩阵的边界,即左上角和右下角 - 下面是Ruby代码。
输入: 一个N*N的二维数组矩阵,矩阵中包含正数和负数元素。
输出: 任意大小的子矩阵,其所有子矩阵中元素之和最大。
要求: 算法复杂度为O(N^3)
历史: 在Algorithmist Larry和Kadane算法的修改帮助下,我已经解决了部分问题,即仅确定总和 - 下面是Java代码。
感谢Ernesto解决了该问题的其余部分,即确定矩阵的边界,即左上角和右下角 - 下面是Ruby代码。
这里是一段解释,用于说明贴出的代码。实现这个问题需要掌握两个关键技巧:(I) Kadane算法和(II)使用前缀和。你还需要将这些技巧应用到矩阵中 (III)。
第一部分:Kadane算法
Kadane算法是一种找到最大连续子序列的方法。让我们从暴力算法开始寻找最大连续子序列,然后考虑优化它以得到Kadane算法。
假设你有一个序列:
-1, 2, 3, -2
对于暴力法,按照下面所示的方式沿着序列走,生成所有可能的子序列。考虑到所有可能性,我们可以在每一步开始、扩展或结束一个列表。
At index 0, we consider appending the -1
-1, 2, 3, -2
^
Possible subsequences:
-1 [sum -1]
At index 1, we consider appending the 2
-1, 2, 3, -2
^
Possible subsequences:
-1 (end) [sum -1]
-1, 2 [sum 1]
2 [sum 2]
At index 2, we consider appending the 3
-1, 2, 3, -2
^
Possible subsequences:
-1, (end) [sum -1]
-1, 2 (end) [sum -1]
2 (end) [sum 2]
-1, 2, 3 [sum 4]
2, 3 [sum 5]
3 [sum 3]
At index 3, we consider appending the -2
-1, 2, 3, -2
^
Possible subsequences:
-1, (end) [sum -1]
-1, 2 (end) [sum 1]
2 (end) [sum 2]
-1, 2 3 (end) [sum 4]
2, 3 (end) [sum 5]
3, (end) [sum 3]
-1, 2, 3, -2 [sum 2]
2, 3, -2 [sum 3]
3, -2 [sum 1]
-2 [sum -2]
对于这种暴力方法,我们最终选择拥有最佳总和的列表(2, 3)
,那就是答案。然而,为了使其更加高效,考虑到你实际上不需要保存每一个列表。在尚未结束的列表中,只需保留最好的一个,其他列表不能做得更好。对于已经结束的列表,只有当其比尚未结束的列表更好时,才需要保留最佳的一个。
因此,你可以通过只使用一个位置数组和一个求和数组来跟踪所需信息。位置数组定义如下:position[r] = s
表示以r
结尾且从s
开始的列表。而sum[r]
给出以索引r
结尾的子序列的总和。这种优化方法称为Kadane算法。
再次运行示例,按照此方式跟踪进度:
At index 0, we consider appending the -1
-1, 2, 3, -2
^
We start a new subsequence for the first element.
position[0] = 0
sum[0] = -1
At index 1, we consider appending the 2
-1, 2, 3, -2
^
We choose to start a new subsequence because that gives a higher sum than extending.
position[0] = 0 sum[0] = -1
position[1] = 1 sum[1] = 2
At index 2, we consider appending the 3
-1, 2, 3, -2
^
We choose to extend a subsequence because that gives a higher sum than starting a new one.
position[0] = 0 sum[0] = -1
position[1] = 1 sum[1] = 2
position[2] = 1 sum[2] = 5
Again, we choose to extend because that gives a higher sum that starting a new one.
-1, 2, 3, -2
^
position[0] = 0 sum[0] = -1
position[1] = 1 sum[1] = 2
position[2] = 1 sum[2] = 5
positions[3] = 3 sum[3] = 3
再次强调,最佳的总和是5,列表从索引1到索引2,即(2, 3)。
第二部分: 前缀和
我们希望有一种方法来计算任意起点到任意终点的行总和。我希望以O(1)时间计算该总和,而不仅仅是加法,其需要O(m)时间,其中m是总和中元素的数量。通过预处理,可以实现这一点。以下是如何实现的。假设你有一个矩阵:
a d g
b e h
c f i
你可以预先计算这个矩阵:
a d g
a+b d+e g+h
a+b+c d+e+f g+h+i
完成这个步骤之后,你可以通过减去两个值,在任意一列中从任何起点到终点计算和。
第三部分:结合技巧寻找最大子矩阵
假设您知道最大子矩阵的顶行和底行。您可以执行以下操作:
那么,如何确定顶行和底行呢? 尝试所有可能性。 尝试将顶部放在您可以的任何位置,并将底部放在您可以的任何位置,并为每个可能性运行先前描述的Kadane基本过程。 找到最大值时,记录顶部和底部位置。
找到行和列需要O(M^2)时间,其中M是行数。 找到列需要O(N)时间,其中N是列数。因此总时间复杂度为O(M^2 * N)。 如果M = N,则所需的时间为O(N ^ 3)。
关于恢复实际子矩阵而不仅仅是最大和,这是我得到的代码。很抱歉我没有时间将我的代码翻译成你的 Java 版本,因此我会在关键部分发布带有注释的 Ruby 代码。
def max_contiguous_submatrix_n3(m)
rows = m.count
cols = rows ? m.first.count : 0
vps = Array.new(rows)
for i in 0..rows
vps[i] = Array.new(cols, 0)
end
for j in 0...cols
vps[0][j] = m[0][j]
for i in 1...rows
vps[i][j] = vps[i-1][j] + m[i][j]
end
end
max = [m[0][0],0,0,0,0] # this is the result, stores [max,top,left,bottom,right]
# these arrays are used over Kadane
sum = Array.new(cols) # obvious sum array used in Kadane
pos = Array.new(cols) # keeps track of the beginning position for the max subseq ending in j
for i in 0...rows
for k in i...rows
# Kadane over all columns with the i..k rows
sum.fill(0) # clean both the sum and pos arrays for the upcoming Kadane
pos.fill(0)
local_max = 0 # we keep track of the position of the max value over each Kadane's execution
# notice that we do not keep track of the max value, but only its position
sum[0] = vps[k][0] - (i==0 ? 0 : vps[i-1][0])
for j in 1...cols
value = vps[k][j] - (i==0 ? 0 : vps[i-1][j])
if sum[j-1] > 0
sum[j] = sum[j-1] + value
pos[j] = pos[j-1]
else
sum[j] = value
pos[j] = j
end
if sum[j] > sum[local_max]
local_max = j
end
end
# Kadane ends here
# Here's the key thing
# If the max value obtained over the past Kadane's execution is larger than
# the current maximum, then update the max array with sum and bounds
if sum[local_max] > max[0]
# sum[local_max] is the new max value
# the corresponding submatrix goes from rows i..k.
# and from columns pos[local_max]..local_max
# the array below contains [max_sum,top,left,bottom,right]
max = [sum[local_max], i, pos[local_max], k, local_max]
end
end
end
return max # return the array with [max_sum,top,left,bottom,right]
end
一些澄清的注释:
我使用一个数组来存储与结果相关的所有值,以便于操作。你也可以使用五个独立的变量:max,top,left,bottom,right。只是将这些值在一行中分配给数组更容易,然后子例程返回具有所有必要信息的数组。
如果你将此代码复制并粘贴到支持Ruby的文本高亮编辑器中,显然你会更好地理解它。希望这有所帮助!
已经有很多答案了,但这是我写的另一个Java实现。它比较了三种解决方案:
这里有n = 10到n = 70的样本运行,每隔10个增加一次,并且有一个漂亮的输出来比较运行时间和空间需求。
代码:
public class MaxSubarray2D {
static int LENGTH;
final static int MAX_VAL = 10;
public static void main(String[] args) {
for (int i = 10; i <= 70; i += 10) {
LENGTH = i;
int[][] a = new int[LENGTH][LENGTH];
for (int row = 0; row < LENGTH; row++) {
for (int col = 0; col < LENGTH; col++) {
a[row][col] = (int) (Math.random() * (MAX_VAL + 1));
if (Math.random() > 0.5D) {
a[row][col] = -a[row][col];
}
//System.out.printf("%4d", a[row][col]);
}
//System.out.println();
}
System.out.println("N = " + LENGTH);
System.out.println("-------");
long start, end;
start = System.currentTimeMillis();
naiveSolution(a);
end = System.currentTimeMillis();
System.out.println(" run time: " + (end - start) + " ms no auxiliary space requirements");
start = System.currentTimeMillis();
dynamicProgammingSolution(a);
end = System.currentTimeMillis();
System.out.println(" run time: " + (end - start) + " ms requires auxiliary space for "
+ ((int) Math.pow(LENGTH, 4)) + " integers");
start = System.currentTimeMillis();
kadane2D(a);
end = System.currentTimeMillis();
System.out.println(" run time: " + (end - start) + " ms requires auxiliary space for " +
+ ((int) Math.pow(LENGTH, 2)) + " integers");
System.out.println();
System.out.println();
}
}
// O(N^2) !!!
public static void kadane2D(int[][] a) {
int[][] s = new int[LENGTH + 1][LENGTH]; // [ending row][sum from row zero to ending row] (rows 1-indexed!)
for (int r = 0; r < LENGTH + 1; r++) {
for (int c = 0; c < LENGTH; c++) {
s[r][c] = 0;
}
}
for (int r = 1; r < LENGTH + 1; r++) {
for (int c = 0; c < LENGTH; c++) {
s[r][c] = s[r - 1][c] + a[r - 1][c];
}
}
int maxSum = Integer.MIN_VALUE;
int maxRowStart = -1;
int maxColStart = -1;
int maxRowEnd = -1;
int maxColEnd = -1;
for (int r1 = 1; r1 < LENGTH + 1; r1++) { // rows 1-indexed!
for (int r2 = r1; r2 < LENGTH + 1; r2++) { // rows 1-indexed!
int[] s1 = new int[LENGTH];
for (int c = 0; c < LENGTH; c++) {
s1[c] = s[r2][c] - s[r1 - 1][c];
}
int max = 0;
int c1 = 0;
for (int c = 0; c < LENGTH; c++) {
max = s1[c] + max;
if (max <= 0) {
max = 0;
c1 = c + 1;
}
if (max > maxSum) {
maxSum = max;
maxRowStart = r1 - 1;
maxColStart = c1;
maxRowEnd = r2 - 1;
maxColEnd = c;
}
}
}
}
System.out.print("KADANE SOLUTION | Max sum: " + maxSum);
System.out.print(" Start: (" + maxRowStart + ", " + maxColStart +
") End: (" + maxRowEnd + ", " + maxColEnd + ")");
}
// O(N^4) !!!
public static void dynamicProgammingSolution(int[][] a) {
int[][][][] dynTable = new int[LENGTH][LENGTH][LENGTH + 1][LENGTH + 1]; // [row][col][height][width]
int maxSum = Integer.MIN_VALUE;
int maxRowStart = -1;
int maxColStart = -1;
int maxRowEnd = -1;
int maxColEnd = -1;
for (int r = 0; r < LENGTH; r++) {
for (int c = 0; c < LENGTH; c++) {
for (int h = 0; h < LENGTH + 1; h++) {
for (int w = 0; w < LENGTH + 1; w++) {
dynTable[r][c][h][w] = 0;
}
}
}
}
for (int r = 0; r < LENGTH; r++) {
for (int c = 0; c < LENGTH; c++) {
for (int h = 1; h <= LENGTH - r; h++) {
int rowTotal = 0;
for (int w = 1; w <= LENGTH - c; w++) {
rowTotal += a[r + h - 1][c + w - 1];
dynTable[r][c][h][w] = rowTotal + dynTable[r][c][h - 1][w];
}
}
}
}
for (int r = 0; r < LENGTH; r++) {
for (int c = 0; c < LENGTH; c++) {
for (int h = 0; h < LENGTH + 1; h++) {
for (int w = 0; w < LENGTH + 1; w++) {
if (dynTable[r][c][h][w] > maxSum) {
maxSum = dynTable[r][c][h][w];
maxRowStart = r;
maxColStart = c;
maxRowEnd = r + h - 1;
maxColEnd = c + w - 1;
}
}
}
}
}
System.out.print(" DP SOLUTION | Max sum: " + maxSum);
System.out.print(" Start: (" + maxRowStart + ", " + maxColStart +
") End: (" + maxRowEnd + ", " + maxColEnd + ")");
}
// O(N^6) !!!
public static void naiveSolution(int[][] a) {
int maxSum = Integer.MIN_VALUE;
int maxRowStart = -1;
int maxColStart = -1;
int maxRowEnd = -1;
int maxColEnd = -1;
for (int rowStart = 0; rowStart < LENGTH; rowStart++) {
for (int colStart = 0; colStart < LENGTH; colStart++) {
for (int rowEnd = 0; rowEnd < LENGTH; rowEnd++) {
for (int colEnd = 0; colEnd < LENGTH; colEnd++) {
int sum = 0;
for (int row = rowStart; row <= rowEnd; row++) {
for (int col = colStart; col <= colEnd; col++) {
sum += a[row][col];
}
}
if (sum > maxSum) {
maxSum = sum;
maxRowStart = rowStart;
maxColStart = colStart;
maxRowEnd = rowEnd;
maxColEnd = colEnd;
}
}
}
}
}
System.out.print(" NAIVE SOLUTION | Max sum: " + maxSum);
System.out.print(" Start: (" + maxRowStart + ", " + maxColStart +
") End: (" + maxRowEnd + ", " + maxColEnd + ")");
}
}
public int[][] findMaximumSubMatrix(int[][] matrix){
int dim = matrix.length;
//computing the vertical prefix sum for columns
int[][] ps = new int[dim][dim];
for (int i = 0; i < dim; i++) {
for (int j = 0; j < dim; j++) {
if (j == 0) {
ps[j][i] = matrix[j][i];
} else {
ps[j][i] = matrix[j][i] + ps[j - 1][i];
}
}
}
int maxSum = matrix[0][0];
int top = 0, left = 0, bottom = 0, right = 0;
//Auxiliary variables
int[] sum = new int[dim];
int[] pos = new int[dim];
int localMax;
for (int i = 0; i < dim; i++) {
for (int k = i; k < dim; k++) {
// Kadane over all columns with the i..k rows
reset(sum);
reset(pos);
localMax = 0;
//we keep track of the position of the max value over each Kadane's execution
// notice that we do not keep track of the max value, but only its position
sum[0] = ps[k][0] - (i==0 ? 0 : ps[i-1][0]);
for (int j = 1; j < dim; j++) {
if (sum[j-1] > 0){
sum[j] = sum[j-1] + ps[k][j] - (i==0 ? 0 : ps[i-1][j]);
pos[j] = pos[j-1];
}else{
sum[j] = ps[k][j] - (i==0 ? 0 : ps[i-1][j]);
pos[j] = j;
}
if (sum[j] > sum[localMax]){
localMax = j;
}
}//Kadane ends here
if (sum[localMax] > maxSum){
/* sum[localMax] is the new max value
the corresponding submatrix goes from rows i..k.
and from columns pos[localMax]..localMax
*/
maxSum = sum[localMax];
top = i;
left = pos[localMax];
bottom = k;
right = localMax;
}
}
}
System.out.println("Max SubMatrix determinant = " + maxSum);
//composing the required matrix
int[][] output = new int[bottom - top + 1][right - left + 1];
for(int i = top, k = 0; i <= bottom; i++, k++){
for(int j = left, l = 0; j <= right ; j++, l++){
output[k][l] = matrix[i][j];
}
}
return output;
}
private void reset(int[] a) {
for (int index = 0; index < a.length; index++) {
a[index] = 0;
}
}
在算法工程师和Larry的帮助下,并使用Kadane算法的修改版本,这是我的解决方案:
int dim = matrix.length;
//computing the vertical prefix sum for columns
int[][] ps = new int[dim][dim];
for (int i = 0; i < dim; i++) {
for (int j = 0; j < dim; j++) {
if (j == 0) {
ps[j][i] = matrix[j][i];
} else {
ps[j][i] = matrix[j][i] + ps[j - 1][i];
}
}
}
int maxSoFar = 0;
int min , subMatrix;
//iterate over the possible combinations applying Kadane's Alg.
for (int i = 0; i < dim; i++) {
for (int j = i; j < dim; j++) {
min = 0;
subMatrix = 0;
for (int k = 0; k < dim; k++) {
if (i == 0) {
subMatrix += ps[j][k];
} else {
subMatrix += ps[j][k] - ps[i - 1 ][k];
}
if(subMatrix < min){
min = subMatrix;
}
if((subMatrix - min) > maxSoFar){
maxSoFar = subMatrix - min;
}
}
}
}
这是我实现的二维Kadane算法。我认为它更加清晰易懂。该算法基于Kadane算法,主要部分的第一和第二个循环用于选择每一行的组合,第三个循环使用一维Kadane算法,通过计算每列的和(由于矩阵的预处理,可在常数时间内完成,通过从组合中选择两行并减去其值)。以下是代码:
int [][] m = {
{1,-5,-5},
{1,3,-5},
{1,3,-5}
};
int N = m.length;
// summing columns to be able to count sum between two rows in some column in const time
for (int i=0; i<N; ++i)
m[0][i] = m[0][i];
for (int j=1; j<N; ++j)
for (int i=0; i<N; ++i)
m[j][i] = m[j][i] + m[j-1][i];
int total_max = 0, sum;
for (int i=0; i<N; ++i) {
for (int k=i; k<N; ++k) { //for each combination of rows
sum = 0;
for (int j=0; j<N; j++) { //kadane algorithm for every column
sum += i==0 ? m[k][j] : m[k][j] - m[i-1][j]; //for first upper row is exception
total_max = Math.max(sum, total_max);
}
}
}
System.out.println(total_max);
O(i^3j^3) -- naive brute force method
o(i^2j^2) -- dynamic programming with memoization
O(i^2j) -- using max contiguous sub sequence for an array
if ( i == j )
O(n^6) -- naive
O(n^4) -- dynamic programming
O(n^3) -- max contiguous sub sequence
这是C#的解决方案。参考:http://www.algorithmist.com/index.php/UVa_108
public static MaxSumMatrix FindMaxSumSubmatrix(int[,] inMtrx)
{
MaxSumMatrix maxSumMtrx = new MaxSumMatrix();
// Step 1. Create SumMatrix - do the cumulative columnar summation
// S[i,j] = S[i-1,j]+ inMtrx[i-1,j];
int m = inMtrx.GetUpperBound(0) + 2;
int n = inMtrx.GetUpperBound(1)+1;
int[,] sumMatrix = new int[m, n];
for (int i = 1; i < m; i++)
{
for (int j = 0; j < n; j++)
{
sumMatrix[i, j] = sumMatrix[i - 1, j] + inMtrx[i - 1, j];
}
}
PrintMatrix(sumMatrix);
// Step 2. Create rowSpans starting each rowIdx. For these row spans, create a 1-D array r_ij
for (int x = 0; x < n; x++)
{
for (int y = x; y < n; y++)
{
int[] r_ij = new int[n];
for (int k = 0; k < n; k++)
{
r_ij[k] = sumMatrix[y + 1,k] - sumMatrix[x, k];
}
// Step 3. Find MaxSubarray of this r_ij. If the sum is greater than the last recorded sum =>
// capture Sum, colStartIdx, ColEndIdx.
// capture current x as rowTopIdx, y as rowBottomIdx.
MaxSum currMaxSum = KadanesAlgo.FindMaxSumSubarray(r_ij);
if (currMaxSum.maxSum > maxSumMtrx.sum)
{
maxSumMtrx.sum = currMaxSum.maxSum;
maxSumMtrx.colStart = currMaxSum.maxStartIdx;
maxSumMtrx.colEnd = currMaxSum.maxEndIdx;
maxSumMtrx.rowStart = x;
maxSumMtrx.rowEnd = y;
}
}
}
return maxSumMtrx;
}
public static void PrintMatrix(int[,] matrix)
{
int endRow = matrix.GetUpperBound(0);
int endCol = matrix.GetUpperBound(1);
PrintMatrix(matrix, 0, endRow, 0, endCol);
}
public static void PrintMatrix(int[,] matrix, int startRow, int endRow, int startCol, int endCol)
{
StringBuilder sb = new StringBuilder();
for (int i = startRow; i <= endRow; i++)
{
sb.Append(Environment.NewLine);
for (int j = startCol; j <= endCol; j++)
{
sb.Append(string.Format("{0} ", matrix[i,j]));
}
}
Console.WriteLine(sb.ToString());
}
// Given an NxN matrix of positive and negative integers, write code to find the sub-matrix with the largest possible sum
public static MaxSum FindMaxSumSubarray(int[] inArr)
{
int currMax = 0;
int currStartIndex = 0;
// initialize maxSum to -infinity, maxStart and maxEnd idx to 0.
MaxSum mx = new MaxSum(int.MinValue, 0, 0);
// travers through the array
for (int currEndIndex = 0; currEndIndex < inArr.Length; currEndIndex++)
{
// add element value to the current max.
currMax += inArr[currEndIndex];
// if current max is more that the last maxSum calculated, set the maxSum and its idx
if (currMax > mx.maxSum)
{
mx.maxSum = currMax;
mx.maxStartIdx = currStartIndex;
mx.maxEndIdx = currEndIndex;
}
if (currMax < 0) // if currMax is -ve, change it back to 0
{
currMax = 0;
currStartIndex = currEndIndex + 1;
}
}
return mx;
}
struct MaxSum
{
public int maxSum;
public int maxStartIdx;
public int maxEndIdx;
public MaxSum(int mxSum, int mxStart, int mxEnd)
{
this.maxSum = mxSum;
this.maxStartIdx = mxStart;
this.maxEndIdx = mxEnd;
}
}
class MaxSumMatrix
{
public int sum = int.MinValue;
public int rowStart = -1;
public int rowEnd = -1;
public int colStart = -1;
public int colEnd = -1;
}
这是我的解决方案。时间复杂度为O(n^3),空间复杂度为O(n^2)。 https://gist.github.com/toliuweijing/6097144
// 0th O(n) on all candidate bottoms @B.
// 1th O(n) on candidate tops @T.
// 2th O(n) on finding the maximum @left/@right match.
int maxRect(vector<vector<int> >& mat) {
int n = mat.size();
vector<vector<int> >& colSum = mat;
for (int i = 1 ; i < n ; ++i)
for (int j = 0 ; j < n ; ++j)
colSum[i][j] += colSum[i-1][j];
int optrect = 0;
for (int b = 0 ; b < n ; ++b) {
for (int t = 0 ; t <= b ; ++t) {
int minLeft = 0;
int rowSum[n];
for (int i = 0 ; i < n ; ++i) {
int col = t == 0 ? colSum[b][i] : colSum[b][i] - colSum[t-1][i];
rowSum[i] = i == 0? col : col + rowSum[i-1];
optrect = max(optrect, rowSum[i] - minLeft);
minLeft = min(minLeft, rowSum[i]);
}
}
}
return optrect;
}
O(N^3)
解决方案的内容。 - Larryn^2
行,这就是你的组合。如果你已经有了部分和,你可以在O(1)
时间内查询下一列(在这些行内),这类似于处理传统的一维Kadane算法中的单个元素。 - Larry