本文概述
给定一个整数数组和一个整数值k, 找出k个最大和为k的非重叠子数组。
例子:
Input : arr1[] = {4, 1, 1, -1, -3, -5, 6, 2, -6, -2}, k = 3.
Output : Maximum non-overlapping sub-array sum1: 8, starting index: 6, ending index: 7.
Maximum non-overlapping sub-array sum2: 6, starting index: 0, ending index: 2.
Maximum non-overlapping sub-array sum3: -1, starting index: 3, ending index: 3.
Input : arr2 = {5, 1, 2, -6, 2, -1, 3, 1}, k = 2.
Output : Maximum non-overlapping sub-array sum1: 8, starting index: 0, ending index: 2.
Maximum non-overlapping sub-array sum2: 5, starting index: 4, ending index: 7.
推荐:请尝试以下方法{IDE}首先, 在继续解决方案之前。
Kadane的算法仅找出最大子阵列和, 但是使用相同的算法, 我们可以找出k个最大非重叠子阵列和。方法是:
- 使用Kadane的算法找出阵列中的最大子阵列。还找出其开始和结束索引。打印此子数组的总和。
- 用-infinity填充此子数组的每个单元格。
- 重复过程1和2 k次。
C ++
// C++ program to find out k maximum
// non-overlapping sub-array sums.
#include <bits/stdc++.h>
using namespace std;
// Function to compute k maximum
// sub-array sums.
void kmax( int arr[], int k, int n) {
// In each iteration it will give
// the ith maximum subarray sum.
for ( int c = 0; c < k; c++){
// Kadane's algorithm.
int max_so_far = numeric_limits< int >::min();
int max_here = 0;
// compute starting and ending
// index of each of the sub-array.
int start = 0, end = 0, s = 0;
for ( int i = 0; i < n; i++)
{
max_here += arr[i];
if (max_so_far < max_here)
{
max_so_far = max_here;
start = s;
end = i;
}
if (max_here < 0)
{
max_here = 0;
s = i + 1;
}
}
// Print out the result.
cout << "Maximum non-overlapping sub-array sum"
<< (c + 1) << ": " << max_so_far
<< ", starting index: " << start
<< ", ending index: " << end << "." << endl;
// Replace all elements of the maximum subarray
// by -infinity. Hence these places cannot form
// maximum sum subarray again.
for ( int l = start; l <= end; l++)
arr[l] = numeric_limits< int >::min();
}
cout << endl;
}
// Driver Program
int main()
{
// Test case 1
int arr1[] = {4, 1, 1, -1, -3, -5, 6, 2, -6, -2};
int k1 = 3;
int n1 = sizeof (arr1) / sizeof (arr1[0]);
// Function calling
kmax(arr1, k1, n1);
// Test case 2
int arr2[] = {5, 1, 2, -6, 2, -1, 3, 1};
int k2 = 2;
int n2 = sizeof (arr2)/ sizeof (arr2[0]);
// Function calling
kmax(arr2, k2, n2);
return 0;
}
Java
// Java program to find out k maximum
// non-overlapping sub-array sums.
class GFG {
// Method to compute k maximum
// sub-array sums.
static void kmax( int arr[], int k, int n) {
// In each iteration it will give
// the ith maximum subarray sum.
for ( int c = 0 ; c < k; c++)
{
// Kadane's algorithm.
int max_so_far = Integer.MIN_VALUE;
int max_here = 0 ;
// compute starting and ending
// index of each of the sub-array.
int start = 0 , end = 0 , s = 0 ;
for ( int i = 0 ; i < n; i++)
{
max_here += arr[i];
if (max_so_far < max_here)
{
max_so_far = max_here;
start = s;
end = i;
}
if (max_here < 0 )
{
max_here = 0 ;
s = i + 1 ;
}
}
// Print out the result.
System.out.println( "Maximum non-overlapping sub-arraysum" +
(c + 1 ) + ": " + max_so_far +
", starting index: " + start +
", ending index: " + end + "." );
// Replace all elements of the maximum subarray
// by -infinity. Hence these places cannot form
// maximum sum subarray again.
for ( int l = start; l <= end; l++)
arr[l] = Integer.MIN_VALUE;
}
System.out.println();
}
// Driver Program
public static void main(String[] args)
{
// Test case 1
int arr1[] = { 4 , 1 , 1 , - 1 , - 3 , - 5 , 6 , 2 , - 6 , - 2 };
int k1 = 3 ;
int n1 = arr1.length;
// Function calling
kmax(arr1, k1, n1);
// Test case 2
int arr2[] = { 5 , 1 , 2 , - 6 , 2 , - 1 , 3 , 1 };
int k2 = 2 ;
int n2 = arr2.length;
// Function calling
kmax(arr2, k2, n2);
}
}
// This code is contributed by Nirmal Patel
Python3
# Python program to find out k maximum
# non-overlapping subarray sums.
# Function to compute k
# maximum sub-array sums.
def kmax(arr, k, n):
# In each iteration it will give
# the ith maximum subarray sum.
for c in range (k):
# Kadane's algorithm
max_so_far = - float ( "inf" )
max_here = 0
# compute starting and ending
# index of each of the subarray
start = 0
end = 0
s = 0
for i in range (n):
max_here + = arr[i]
if (max_so_far < max_here):
max_so_far = max_here
start = s
end = i
if (max_here < 0 ):
max_here = 0
s = i + 1
# Print out the result
print ( "Maximum non-overlapping sub-array sum" , c + 1 , ": " , max_so_far, ", starting index: " , start, ", ending index: " , end, "." , sep = "")
# Replace all elements of the maximum subarray
# by -infinity. Hence these places cannot form
# maximum sum subarray again.
for l in range (start, end + 1 ):
arr[l] = - float ( "inf" )
print ()
# Driver Program
# Test case 1
arr1 = [ 4 , 1 , 1 , - 1 , - 3 , - 5 , 6 , 2 , - 6 , - 2 ]
k1 = 3
n1 = len (arr1)
# Function calling
kmax(arr1, k1, n1)
# Test case 2
arr2 = [ 5 , 1 , 2 , - 6 , 2 , - 1 , 3 , 1 ]
k2 = 2
n2 = len (arr2)
# Function calling
kmax(arr2, k2, n2)
C#
// C# program to find out k maximum
// non-overlapping sub-array sums.
using System;
class GFG {
// Method to compute k
// maximum sub-array sums.
static void kmax( int []arr, int k, int n) {
// In each iteration it will give
// the ith maximum subarray sum.
for ( int c = 0; c < k; c++)
{
// Kadane's algorithm.
int max_so_far = int .MinValue;
int max_here = 0;
// compute starting and ending
// index of each of the sub-array.
int start = 0, end = 0, s = 0;
for ( int i = 0; i < n; i++)
{
max_here += arr[i];
if (max_so_far < max_here)
{
max_so_far = max_here;
start = s;
end = i;
}
if (max_here < 0)
{
max_here = 0;
s = i + 1;
}
}
// Print out the result.
Console.WriteLine( "Maximum non-overlapping sub-arraysum" +
(c + 1) + ": " + max_so_far +
", starting index: " + start +
", ending index: " + end + "." );
// Replace all elements of the maximum subarray
// by -infinity. Hence these places cannot form
// maximum sum subarray again.
for ( int l = start; l <= end; l++)
arr[l] = int .MinValue;
}
Console.WriteLine();
}
// Driver Program
public static void Main(String[] args)
{
// Test case 1
int []arr1 = {4, 1, 1, -1, -3, -5, 6, 2, -6, -2};
int k1 = 3;
int n1 = arr1.Length;
// Function calling
kmax(arr1, k1, n1);
// Test case 2
int []arr2 = {5, 1, 2, -6, 2, -1, 3, 1};
int k2 = 2;
int n2 = arr2.Length;
// Function calling
kmax(arr2, k2, n2);
}
}
// This code is contributed by parashar...
的PHP
<?php
// PHP program to find out k maximum
// non-overlapping sub-array sums.
// Method to compute k
// maximum sub-array sums.
function kmax( $arr , $k , $n ) {
// In each iteration it will give
// the ith maximum subarray sum.
for ( $c = 0; $c < $k ; $c ++)
{
// Kadane's algorithm.
$max_so_far = PHP_INT_MIN;
$max_here = 0;
// compute starting and ending
// index of each of the sub-array.
$start = 0; $end = 0; $s = 0;
for ( $i = 0; $i < $n ; $i ++)
{
$max_here += $arr [ $i ];
if ( $max_so_far < $max_here )
{
$max_so_far = $max_here ;
$start = $s ;
$end = $i ;
}
if ( $max_here < 0)
{
$max_here = 0;
$s = $i + 1;
}
}
// Print out the result.
echo "Maximum non-overlapping sub-arraysum" ;
echo ( $c + 1) , ": " , $max_so_far ;
echo ", starting index: " , $start ;
echo ", ending index: " , $end , "." ;
echo "\n" ;
// Replace all elements of the maximum subarray
// by -infinity. Hence these places cannot form
// maximum sum subarray again.
for ( $l = $start ; $l <= $end ; $l ++)
$arr [ $l ] = PHP_INT_MIN;
}
echo "\n" ;
}
// Driver Program
// Test case 1
$arr1 = array (4, 1, 1, -1, -3, -5, 6, 2, -6, -2);
$k1 = 3;
$n1 = count ( $arr1 );
// Function calling
kmax( $arr1 , $k1 , $n1 );
// Test case 2
$arr2 = array (5, 1, 2, -6, 2, -1, 3, 1);
$k2 = 2;
$n2 = count ( $arr2 );
// Function calling
kmax( $arr2 , $k2 , $n2 );
// This code is contributed by anuj_67.
?>
输出如下:
Maximum non-overlapping sub-array sum1: 8, starting index: 6, ending index: 7.
Maximum non-overlapping sub-array sum2: 6, starting index: 0, ending index: 2.
Maximum non-overlapping sub-array sum3: -1, starting index: 3, ending index: 3.
Maximum non-overlapping sub-array sum1: 8, starting index: 0, ending index: 2.
Maximum non-overlapping sub-array sum2: 5, starting index: 4, ending index: 7.
时间复杂度:
外循环运行k次, kadane的算法在每次迭代中以线性时间O(n)运行。因此, 总的时间复杂度为O(k * n)。