算法題：重疊的連續子數組的K個最大和

給定一個整數數組和一個整數值k, 找出k個子數組(可能重疊), 它們具備k個最大和。

例子：

Input : arr = {4, -8, 9, -4, 1, -8, -1, 6}, k = 4
Output : 9 6 6 5


Input : arr = {-2, -3, 4, -1, -2, 1, 5, -3}, k= 3
Output : 7 6 5

使用Kadane的算法咱們能夠找到一個數組的最大連續子數組總和。可是在這種狀況下Kadane的算法不起做用。正如咱們在數組中命中負數時同樣, 將max_ending_here變量設置爲零, 所以咱們錯過了第二個最大值的可能性。

這是咱們提出的一種算法宋恩培和高岡忠雄計算最大子陣列和問題在O(n)時間中, k個最大子數組和問題在O(k * n)時間中。

首先, 咱們使用這種方法研究僅最大子數組和的問題：

先決條件：

1. 前綴和數組
2. O(n)中使用前綴和的最大子數組和

k個最大子數組的方法：

1. Calculate the prefix sum of the input array.
2. Take cand, maxi and mini as arrays of size k. 
3. Initialize mini[0] = 0 for the same reason as previous.
4. for each value of the prefix_sum[i] do
       (i). update cand[j] value by prefix_sum[i] - mini[j]
       (ii). maxi will be the maximum k elements of maxi and cand
       (iii). if prefix_sum is less than all values of mini, then 
              include it in mini and remove the maximum element from mini
       // After the ith iteration mini holds k minimum prefix sum upto
       // index i and maxi holds k maximum overlapping sub-array sums 
       // upto index i.
5. return maxi

在整個計算方法中, 咱們將maxi保持不變, 而mini則保持不變。

C ++

// C++ program to find out k maximum sum of
// overlapping sub-arrays
#include <iostream>
#include <limits>
#include <vector>
  
using namespace std;
  
// Function to compute prefix-sum of the input array
vector< int > prefix_sum(vector< int > arr, int n)
{
     vector< int > pre_sum;
     pre_sum.push_back(arr[0]);
     for ( int i = 1; i < n; i++) 
         pre_sum.push_back(pre_sum[i - 1] + arr[i]);    
     return pre_sum;
}
  
// Update maxi by k maximum values from maxi and cand
void maxMerge(vector< int >& maxi, vector< int > cand)
{
     // Here cand and maxi arrays are in non-increasing
     // order beforehand. Now, j is the index of the
     // next cand element and i is the index of next
     // maxi element. Traverse through maxi array.
     // If cand[j] > maxi[i] insert cand[j] at the ith
     // position in the maxi array and remove the minimum
     // element of the maxi array i.e. the last element
     // and increase j by 1 i.e. take the next element
     // from cand.
     int k = maxi.size();
     int j = 0;
     for ( int i = 0; i < k; i++) {
         if (cand[j] > maxi[i]) {
             maxi.insert(maxi.begin() + i, cand[j]);
             maxi.erase(maxi.begin() + k);
             j += 1;
         }
     }
}
  
// Insert prefix_sum[i] to mini array if needed
void insertMini(vector< int >& mini, int pre_sum)
{
     // Traverse the mini array from left to right.
     // If prefix_sum[i] is less than any element
     // then insert prefix_sum[i] at that position
     // and delete maximum element of the mini array
     // i.e. the rightmost element from the array.
     int k = mini.size();
     for ( int i = 0; i < k; i++) {
         if (pre_sum < mini[i]) {
             mini.insert(mini.begin() + i, pre_sum);
             mini.erase(mini.begin() + k);
             break ;
         }
     }
}
  
// Function to compute k maximum overlapping sub-
// array sums
void kMaxOvSubArray(vector< int > arr, int k)
{
     int n = arr.size();
  
     // Compute the prefix sum of the input array.
     vector< int > pre_sum = prefix_sum(arr, n);
  
     // Set all the elements of mini as +infinite
     // except 0th. Set the 0th element as 0.
     vector< int > mini(k, numeric_limits< int >::max());
     mini[0] = 0;
  
     // Set all the elements of maxi as -infinite.
     vector< int > maxi(k, numeric_limits< int >::min());
  
     // Initialize cand array.
     vector< int > cand(k);
  
     // For each element of the prefix_sum array do:
     // compute the cand array.
     // take k maximum values from maxi and cand
     // using maxmerge function.
     // insert prefix_sum[i] to mini array if needed
     // using insertMini function.
     for ( int i = 0; i < n; i++) {
         for ( int j = 0; j < k; j++) {
              if (pre_sum[i] < 0 && mini[j]==numeric_limits< int >::max())
            cand[j]=(-pre_sum[i])-mini[j];
          else cand[j] = pre_sum[i] - mini[j];
         }
         maxMerge(maxi, cand);
         insertMini(mini, pre_sum[i]);
     }
  
     // Elements of maxi array is the k
     // maximum overlapping sub-array sums.
     // Print out the elements of maxi array.
     for ( int ele : maxi)
         cout << ele << " " ;
     cout << endl;
}
  
// Driver Program
int main()
{
     // Test case 1
     vector< int > arr1 = { 4, -8, 9, -4, 1, -8, -1, 6 };
     int k1 = 4;
     kMaxOvSubArray(arr1, k1);
  
     // Test case 2
     vector< int > arr2 = { -2, -3, 4, -1, -2, 1, 5, -3 };
     int k2 = 3;
     kMaxOvSubArray(arr2, k2);
     return 0;
}

Python3

# Python program to find out k maximum sum of
# overlapping sub-arrays
  
# Function to compute prefix-sum of the input array
def prefix_sum(arr, n):
     pre_sum = list ()
     pre_sum.append(arr[ 0 ])
     for i in range ( 1 , n):
         pre_sum.append(pre_sum[i - 1 ] + arr[i])
     return pre_sum
  
# Update maxi by k maximum values from maxi and cand
def maxMerge(maxi, cand):
  
     # Here cand and maxi arrays are in non-increasing
     # order beforehand. Now, j is the index of the
     # next cand element and i is the index of next
     # maxi element. Traverse through maxi array.
     # If cand[j] > maxi[i] insert cand[j] at the ith
     # position in the maxi array and remove the minimum
     # element of the maxi array i.e. the last element
     # and increase j by 1 i.e. take the next element
     # from cand.
     k = len (maxi)
     j = 0
     for i in range (k):
         if (cand[j] > maxi[i]):
             maxi.insert(i, cand[j])
             del maxi[ - 1 ]
             j + = 1
  
# Insert prefix_sum[i] to mini array if needed
def insertMini(mini, pre_sum):
  
     # Traverse the mini array from left to right.
     # If prefix_sum[i] is less than any element
     # then insert prefix_sum[i] at that position
     # and delete maximum element of the mini array
     # i.e. the rightmost element from the array.
     k = len (mini)
     for i in range (k):
         if (pre_sum < mini[i]):
             mini.insert(i, pre_sum)
             del mini[ - 1 ]
             break
  
# Function to compute k maximum overlapping sub-array sums
def kMaxOvSubArray(arr, k):
     n = len (arr)
  
     # Compute the prefix sum of the input array.
     pre_sum = prefix_sum(arr, n)
  
     # Set all the elements of mini as + infinite
     # except 0th. Set the 0th element as 0.
     mini = [ float ( 'inf' ) for i in range (k)]
     mini[ 0 ] = 0
  
     # Set all the elements of maxi as -infinite.
     maxi = [ - float ( 'inf' ) for i in range (k)]
  
     # Initialize cand array.
     cand = [ 0 for i in range (k)]
  
     # For each element of the prefix_sum array do:
     # compute the cand array.
     # take k maximum values from maxi and cand
     # using maxmerge function.
     # insert prefix_sum[i] to mini array if needed
     # using insertMini function.
     for i in range (n):
         for j in range (k):
             cand[j] = pre_sum[i] - mini[j]
         maxMerge(maxi, cand)
         insertMini(mini, pre_sum[i])
  
     # Elements of maxi array is the k
     # maximum overlapping sub-array sums.
     # Print out the elements of maxi array.
     for ele in maxi:
         print (ele, end = ' ' )
     print ('')
  
# Driver Program
# Test case 1
arr1 = [ 4 , - 8 , 9 , - 4 , 1 , - 8 , - 1 , 6 ]
k1 = 4
kMaxOvSubArray(arr1, k1)
  
# Test case 2
arr2 = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ]
k2 = 3
kMaxOvSubArray(arr2, k2)

輸出以下：

9 6 6 5
7 6 5

時間複雜度：" insertMini"和" maxMerge"函數的運行時間爲O(k), 而更新" cand"數組則須要O(k)時間。咱們執行此過程n次。所以, 總體時間複雜度爲O(k * n).

更多數據結構和算法相關內容請參考：lsbin - IT開發技術：https://www.lsbin.com/

查看如下更多算法相關的內容：

• FCFS CPU調度算法程序實現：https://www.lsbin.com/3839.html
• 如何從列標題中查找Excel列號？：https://www.lsbin.com/3737.html
• 如何計算兩個數組中的最大求和路徑？：https://www.lsbin.com/3689.html