给定大小为N * M的2D矩阵。任务是找到从(0，0)到(N-1，M-1)的最大乘积路径。您只能从(i，j)移至(i，j + 1)并从(i，j)移至(i + 1，j) 。
例子
Input: arr[][] = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} }
Output: 2016
The path with maximum product is : 1->4->7->8->9 = 2016
方法：
要从当前位置选择哪个方向，我们必须检查给出最大乘积的方向。我们将维护两个2D阵列：
- maxPath [i] [j]：它将存储最大乘积，直到(i，j)
- minPath [i] [j]：它将存储最小乘积，直到(i，j)
脚步：
- 将maxPath [0] [0]和minPath [0] [0]初始化为arr [0] [0]。
- 对于maxPath [i] [j]中所有剩余的像元，通过以下方法检查当前像元(i，j)与上一个像元(i-1，j)的乘积是否最大：
considering rows:
maxValue1 = max(arr[i][j]*maxPath[i-1][j], arr[i][j]*minPath[i-1][j])
considering columns:
maxValue2 = max(maxPath[i][j – 1] * arr[i][j], minPath[i][j – 1] * arr[i][j])
as arr[i][j] can be negative and if minPath[i][j] is negative. Then,
arr[i][j]*minPath[i][j] is positive, which can be maximum product. - 对于minPath [i] [j]中所有剩余的像元，通过以下方法检查当前像元(i，j)与前一个像元(i-1，j)的乘积是否最小：
considering rows:
minValue1 = min(maxPath[i – 1][j] * arr[i][j], minPath[i – 1][j] * arr[i][j])
considering columns:
minValue2 = min(maxPath[i][j – 1] * arr[i][j], minPath[i][j – 1] * arr[i][j]) - 更新minPath [i] [j] = min(minValue1，minValue2)和maxPath [i] [j] = max(maxValue1，maxValue2)
- 返回maxPath [n-1] [m-1]以获取最大乘积。
下面是上述方法的实现：
C++
// C++ Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)
#include
using namespace std;
#define N 3
#define M 3
// Function to find maximum product
int maxProductPath(int arr[N][M])
{
// It will store the maximum
// product till a given cell.
vector > maxPath(N, vector(M, INT_MIN));
// It will store the minimum
// product till a given cell
// (for -ve elements)
vector > minPath(N, vector(M, INT_MAX));
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
int minVal = INT_MAX;
int maxVal = INT_MIN;
// If we are at topmost
// or leftmost, just
// copy the elements.
if (i == 0 && j == 0) {
maxVal = arr[i][j];
minVal = arr[i][j];
}
// If we're not at the
// above, we can consider
// the above value.
if (i > 0) {
int tempMax = max(maxPath[i - 1][j] * arr[i][j],
minPath[i - 1][j] * arr[i][j]);
maxVal = max(maxVal, tempMax);
int tempMin = min(maxPath[i - 1][j] * arr[i][j],
minPath[i - 1][j] * arr[i][j]);
minVal = min(minVal, tempMin);
}
// If we're not on the
// leftmost, we can consider
// the left value.
if (j > 0) {
int tempMax = max(maxPath[i][j - 1] * arr[i][j],
minPath[i][j - 1] * arr[i][j]);
maxVal = max(maxVal, tempMax);
int tempMin = min(maxPath[i][j - 1] * arr[i][j],
minPath[i][j - 1] * arr[i][j]);
minVal = min(minVal, tempMin);
}
// Store max & min product
// till i, j.
maxPath[i][j] = maxVal;
minPath[i][j] = minVal;
}
}
// Return the max product path
// from 0, 0 -> N-1, M-1.
return maxPath[N - 1][M - 1];
}
// Driver Code
int main()
{
int arr[N][M] = { { 1, -2, 3 },
{ 4, -5, 6 },
{ -7, -8, 9 } };
// Print maximum product from
// (0, 0) to (N-1, M-1)
cout << maxProductPath(arr) << endl;
return 0;
} Java
// Java Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)
class GFG
{
static final int N = 3;
static final int M = 3;
// Function to find maximum product
static int maxProductPath(int arr[][])
{
// It will store the maximum
// product till a given cell.
int [][]maxPath = new int[N][M];
// It will store the minimum
// product till a given cell
// (for -ve elements)
int [][]minPath = new int[N][M];
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
int minVal = Integer.MAX_VALUE;
int maxVal = Integer.MIN_VALUE;
// If we are at topmost
// or leftmost, just
// copy the elements.
if (i == 0 && j == 0)
{
maxVal = arr[i][j];
minVal = arr[i][j];
}
// If we're not at the
// above, we can consider
// the above value.
if (i > 0)
{
int tempMax = Math.max(maxPath[i - 1][j] * arr[i][j],
minPath[i - 1][j] * arr[i][j]);
maxVal = Math.max(maxVal, tempMax);
int tempMin = Math.min(maxPath[i - 1][j] * arr[i][j],
minPath[i - 1][j] * arr[i][j]);
minVal = Math.min(minVal, tempMin);
}
// If we're not on the
// leftmost, we can consider
// the left value.
if (j > 0) {
int tempMax = Math.max(maxPath[i][j - 1] * arr[i][j],
minPath[i][j - 1] * arr[i][j]);
maxVal = Math.max(maxVal, tempMax);
int tempMin = Math.min(maxPath[i][j - 1] * arr[i][j],
minPath[i][j - 1] * arr[i][j]);
minVal = Math.min(minVal, tempMin);
}
// Store max & min product
// till i, j.
maxPath[i][j] = maxVal;
minPath[i][j] = minVal;
}
}
// Return the max product path
// from 0, 0.N-1, M-1.
return maxPath[N - 1][M - 1];
}
// Driver Code
public static void main(String[] args)
{
int arr[][] = { { 1, -2, 3 },
{ 4, -5, 6 },
{ -7, -8, 9 } };
// Print maximum product from
// (0, 0) to (N-1, M-1)
System.out.print(maxProductPath(arr) +"\n");
}
}
// This code is contributed by 29AjayKumarPython3
# Python Program to find maximum
# product path from (0, 0) to
# (N-1, M-1)
import sys
N = 3;
M = 3;
# Function to find maximum product
def maxProductPath(arr):
# It will store the maximum
# product till a given cell.
maxPath = [[0 for i in range(M)] for j in range(N)];
# It will store the minimum
# product till a given cell
# (for -ve elements)
minPath = [[0 for i in range(M)] for j in range(N)];
for i in range(N):
for j in range(M):
minVal = sys.maxsize;
maxVal = -sys.maxsize;
# If we are at topmost
# or leftmost, just
# copy the elements.
if (i == 0 and j == 0):
maxVal = arr[i][j];
minVal = arr[i][j];
# If we're not at the
# above, we can consider
# the above value.
if (i > 0):
tempMax = max(maxPath[i - 1][j] * arr[i][j],\
minPath[i - 1][j] * arr[i][j]);
maxVal = max(maxVal, tempMax);
tempMin = min(maxPath[i - 1][j] * arr[i][j],\
minPath[i - 1][j] * arr[i][j]);
minVal = min(minVal, tempMin);
# If we're not on the
# leftmost, we can consider
# the left value.
if (j > 0):
tempMax = max(maxPath[i][j - 1] * arr[i][j],\
minPath[i][j - 1] * arr[i][j]);
maxVal = max(maxVal, tempMax);
tempMin = min(maxPath[i][j - 1] * arr[i][j],\
minPath[i][j - 1] * arr[i][j]);
minVal = min(minVal, tempMin);
# Store max & min product
# till i, j.
maxPath[i][j] = maxVal;
minPath[i][j] = minVal;
# Return the max product path
# from 0, 0.N-1, M-1.
return maxPath[N - 1][M - 1];
# Driver Code
if __name__ == '__main__':
arr = [[ 1, -2, 3 ],[ 4, -5, 6 ],[ -7, -8, 9]];
# Prmaximum product from
# (0, 0) to (N-1, M-1)
print(maxProductPath(arr));
# This code is contributed by 29AjayKumarC#
// C# Program to find maximum
// product path from (0, 0) to
// (N-1, M-1)
using System;
class GFG
{
static readonly int N = 3;
static readonly int M = 3;
// Function to find maximum product
static int maxProductPath(int [,]arr)
{
// It will store the maximum
// product till a given cell.
int [,]maxPath = new int[N, M];
// It will store the minimum
// product till a given cell
// (for -ve elements)
int [,]minPath = new int[N, M];
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
int minVal = int.MaxValue;
int maxVal = int.MinValue;
// If we are at topmost
// or leftmost, just
// copy the elements.
if (i == 0 && j == 0)
{
maxVal = arr[i, j];
minVal = arr[i, j];
}
// If we're not at the
// above, we can consider
// the above value.
if (i > 0)
{
int tempMax = Math.Max(maxPath[i - 1,j] * arr[i,j],
minPath[i - 1,j] * arr[i,j]);
maxVal = Math.Max(maxVal, tempMax);
int tempMin = Math.Min(maxPath[i - 1,j] * arr[i,j],
minPath[i - 1,j] * arr[i,j]);
minVal = Math.Min(minVal, tempMin);
}
// If we're not on the
// leftmost, we can consider
// the left value.
if (j > 0) {
int tempMax = Math.Max(maxPath[i,j - 1] * arr[i,j],
minPath[i,j - 1] * arr[i,j]);
maxVal = Math.Max(maxVal, tempMax);
int tempMin = Math.Min(maxPath[i,j - 1] * arr[i,j],
minPath[i,j - 1] * arr[i,j]);
minVal = Math.Min(minVal, tempMin);
}
// Store max & min product
// till i, j.
maxPath[i, j] = maxVal;
minPath[i, j] = minVal;
}
}
// Return the max product path
// from 0, 0.N - 1, M - 1.
return maxPath[N - 1, M - 1];
}
// Driver Code
public static void Main(String[] args)
{
int [,]arr = { { 1, -2, 3 },
{ 4, -5, 6 },
{ -7, -8, 9 } };
// Print maximum product from
// (0, 0) to (N - 1, M - 1)
Console.Write(maxProductPath(arr) +"\n");
}
}
// This code is contributed by 29AjayKumar输出：
2016
时间复杂度： O(N * M)
辅助空间： O(N * M)