本文概述
给定n个数字(+ ve和-ve), 它们排列成一个圆圈, 找出连续数字的最大和。
例子:
Input: a[] = {8, -8, 9, -9, 10, -11, 12}
Output: 22 (12 + 8 - 8 + 9 - 9 + 10)
Input: a[] = {10, -3, -4, 7, 6, 5, -4, -1}
Output: 23 (7 + 6 + 5 - 4 -1 + 10)
Input: a[] = {-1, 40, -14, 7, 6, 5, -4, -1}
Output: 52 (7 + 6 + 5 - 4 - 1 - 1 + 40)
方法1最大和可以有两种情况:
- 情况1:排列有助于最大和的元素, 以使不存在任何包装。示例:{-10、2, -1、5}, {-2、4, -1、4, -1}。在这种情况下, Kadane的算法将产生结果。
- 情况2:布置有助于最大和的元素, 使得存在包装。示例:{10, -12、11}, {12, -5、4, -8、11}。在这种情况下, 我们将换行更改为不换行。让我们看看如何。包装有贡献的元素意味着不包装无贡献的元素, 因此请找出无贡献元素的总和, 然后从总和中减去该总和。要找出无用的总和, 请反转每个元素的符号, 然后运行Kadane的算法。
我们的阵列就像一个圆环, 我们必须消除最大连续负值, 这意味着倒置阵列中的最大连续正值。最后, 我们比较两种情况下获得的总和, 并返回两个总和的最大值。
以下是上述方法的C++/ C++, Java和Python实现。
C ++
//C++ program for maximum contiguous circular sum problem
#include <bits/stdc++.h>
using namespace std;
//Standard Kadane's algorithm to
//find maximum subarray sum
int kadane( int a[], int n);
//The function returns maximum
//circular contiguous sum in a[]
int maxCircularSum( int a[], int n)
{
//Case 1: get the maximum sum using standard kadane'
//s algorithm
int max_kadane = kadane(a, n);
//Case 2: Now find the maximum sum that includes
//corner elements.
int max_wrap = 0, i;
for (i = 0; i <n; i++) {
max_wrap += a[i]; //Calculate array-sum
a[i] = -a[i]; //invert the array (change sign)
}
//max sum with corner elements will be:
//array-sum - (-max subarray sum of inverted array)
max_wrap = max_wrap + kadane(a, n);
//The maximum circular sum will be maximum of two sums
return (max_wrap> max_kadane) ? max_wrap : max_kadane;
}
//Standard Kadane's algorithm to find maximum subarray sum
//See https://www.lsbin.org/archives/576 for details
int kadane( int a[], int n)
{
int max_so_far = 0, max_ending_here = 0;
int i;
for (i = 0; i <n; i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here <0)
max_ending_here = 0;
if (max_so_far <max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
/* Driver program to test maxCircularSum() */
int main()
{
int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
int n = sizeof (a) /sizeof (a[0]);
cout <<"Maximum circular sum is " <<maxCircularSum(a, n) <<endl;
return 0;
}
//This is code is contributed by rathbhupendra
C
//C program for maximum contiguous circular sum problem
#include <stdio.h>
//Standard Kadane's algorithm to find maximum subarray
//sum
int kadane( int a[], int n);
//The function returns maximum circular contiguous sum
//in a[]
int maxCircularSum( int a[], int n)
{
//Case 1: get the maximum sum using standard kadane'
//s algorithm
int max_kadane = kadane(a, n);
//Case 2: Now find the maximum sum that includes
//corner elements.
int max_wrap = 0, i;
for (i = 0; i <n; i++) {
max_wrap += a[i]; //Calculate array-sum
a[i] = -a[i]; //invert the array (change sign)
}
//max sum with corner elements will be:
//array-sum - (-max subarray sum of inverted array)
max_wrap = max_wrap + kadane(a, n);
//The maximum circular sum will be maximum of two sums
return (max_wrap> max_kadane) ? max_wrap : max_kadane;
}
//Standard Kadane's algorithm to find maximum subarray sum
//See https://www.lsbin.org/archives/576 for details
int kadane( int a[], int n)
{
int max_so_far = 0, max_ending_here = 0;
int i;
for (i = 0; i <n; i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here <0)
max_ending_here = 0;
if (max_so_far <max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
/* Driver program to test maxCircularSum() */
int main()
{
int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
int n = sizeof (a) /sizeof (a[0]);
printf ( "Maximum circular sum is %dn" , maxCircularSum(a, n));
return 0;
}
Java
//Java program for maximum contiguous circular sum problem
import java.io.*;
import java.util.*;
class MaxCircularSum {
//The function returns maximum circular contiguous sum
//in a[]
static int maxCircularSum( int a[])
{
int n = a.length;
//Case 1: get the maximum sum using standard kadane'
//s algorithm
int max_kadane = kadane(a);
//Case 2: Now find the maximum sum that includes
//corner elements.
int max_wrap = 0 ;
for ( int i = 0 ; i <n; i++) {
max_wrap += a[i]; //Calculate array-sum
a[i] = -a[i]; //invert the array (change sign)
}
//max sum with corner elements will be:
//array-sum - (-max subarray sum of inverted array)
max_wrap = max_wrap + kadane(a);
//The maximum circular sum will be maximum of two sums
return (max_wrap> max_kadane) ? max_wrap : max_kadane;
}
//Standard Kadane's algorithm to find maximum subarray sum
//See https://www.lsbin.org/archives/576 for details
static int kadane( int a[])
{
int n = a.length;
int max_so_far = 0 , max_ending_here = 0 ;
for ( int i = 0 ; i <n; i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here <0 )
max_ending_here = 0 ;
if (max_so_far <max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
public static void main(String[] args)
{
int a[] = { 11 , 10 , - 20 , 5 , - 3 , - 5 , 8 , - 13 , 10 };
System.out.println( "Maximum circular sum is " + maxCircularSum(a));
}
} /* This code is contributed by Devesh Agrawal*/
python
# Python program for maximum contiguous circular sum problem
# Standard Kadane's algorithm to find maximum subarray sum
def kadane(a):
n = len (a)
max_so_far = 0
max_ending_here = 0
for i in range ( 0 , n):
max_ending_here = max_ending_here + a[i]
if (max_ending_here <0 ):
max_ending_here = 0
if (max_so_far <max_ending_here):
max_so_far = max_ending_here
return max_so_far
# The function returns maximum circular contiguous sum in
# a[]
def maxCircularSum(a):
n = len (a)
# Case 1: get the maximum sum using standard kadane's
# algorithm
max_kadane = kadane(a)
# Case 2: Now find the maximum sum that includes corner
# elements.
max_wrap = 0
for i in range ( 0 , n):
max_wrap + = a[i]
a[i] = - a[i]
# Max sum with corner elements will be:
# array-sum - (-max subarray sum of inverted array)
max_wrap = max_wrap + kadane(a)
# The maximum circular sum will be maximum of two sums
if max_wrap> max_kadane:
return max_wrap
else :
return max_kadane
# Driver function to test above function
a = [ 11 , 10 , - 20 , 5 , - 3 , - 5 , 8 , - 13 , 10 ]
print "Maximum circular sum is" , maxCircularSum(a)
# This code is contributed by Devesh Agrawal
C#
//C# program for maximum contiguous
//circular sum problem
using System;
class MaxCircularSum {
//The function returns maximum circular
//contiguous sum in a[]
static int maxCircularSum( int [] a)
{
int n = a.Length;
//Case 1: get the maximum sum using standard kadane'
//s algorithm
int max_kadane = kadane(a);
//Case 2: Now find the maximum sum that includes
//corner elements.
int max_wrap = 0;
for ( int i = 0; i <n; i++) {
max_wrap += a[i]; //Calculate array-sum
a[i] = -a[i]; //invert the array (change sign)
}
//max sum with corner elements will be:
//array-sum - (-max subarray sum of inverted array)
max_wrap = max_wrap + kadane(a);
//The maximum circular sum will be maximum of two sums
return (max_wrap> max_kadane) ? max_wrap : max_kadane;
}
//Standard Kadane's algorithm to find maximum subarray sum
//See https://www.lsbin.org/archives/576 for details
static int kadane( int [] a)
{
int n = a.Length;
int max_so_far = 0, max_ending_here = 0;
for ( int i = 0; i <n; i++) {
max_ending_here = max_ending_here + a[i];
if (max_ending_here <0)
max_ending_here = 0;
if (max_so_far <max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
//Driver code
public static void Main()
{
int [] a = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
Console.Write( "Maximum circular sum is " + maxCircularSum(a));
}
}
/* This code is contributed by vt_m*/
PHP
<?php
//PHP program for maximum
//contiguous circular sum problem
//The function returns maximum
//circular contiguous sum $a[]
function maxCircularSum( $a , $n )
{
//Case 1: get the maximum sum
//using standard kadane' s algorithm
$max_kadane = kadane( $a , $n );
//Case 2: Now find the maximum
//sum that includes corner elements.
$max_wrap = 0;
for ( $i = 0; $i <$n ; $i ++)
{
$max_wrap += $a [ $i ]; //Calculate array-sum
$a [ $i ] = - $a [ $i ]; //invert the array (change sign)
}
//max sum with corner elements will be:
//array-sum - (-max subarray sum of inverted array)
$max_wrap = $max_wrap + kadane( $a , $n );
//The maximum circular sum will be maximum of two sums
return ( $max_wrap> $max_kadane )? $max_wrap : $max_kadane ;
}
//Standard Kadane's algorithm to
//find maximum subarray sum
//See https://www.lsbin.org/archives/576 for details
function kadane( $a , $n )
{
$max_so_far = 0;
$max_ending_here = 0;
for ( $i = 0; $i <$n ; $i ++)
{
$max_ending_here = $max_ending_here + $a [ $i ];
if ( $max_ending_here <0)
$max_ending_here = 0;
if ( $max_so_far <$max_ending_here )
$max_so_far = $max_ending_here ;
}
return $max_so_far ;
}
/* Driver code */
$a = array (11, 10, -20, 5, -3, -5, 8, -13, 10);
$n = count ( $a );
echo "Maximum circular sum is " . maxCircularSum( $a , $n );
//This code is contributed by rathbhupendra
?>
输出如下:
Maximum circular sum is 31
复杂度分析:
- 时间复杂度:O(n), 其中n是输入数组中元素的数量。
由于仅需要数组的线性遍历。 - 辅助空间:O(1)。
由于不需要额外的空间。
注意如果所有数字均为负数, 例如{-1, -2, -3}, 则上述算法无效。在这种情况下, 它返回0。在运行上述算法之前, 可以通过添加预检查以查看所有数字是否均为负来处理这种情况。
方法2
方法:
在这种方法中, 修改Kadane的算法以找到最小连续子数组和和最大连续子数组和, 然后检查max_value和从总和中减去min_value后剩下的值中的最大值。
算法
- 我们将计算给定数组的总和。
- 我们将变量curr_max, max_so_far, curr_min, min_so_far声明为数组的第一个值。
- 现在, 我们将使用Kadane的算法来找到最大子数组和和最小子数组和。
- 检查数组中的所有值:
- 如果min_so_far等于和, 即所有值均为负, 则返回max_so_far。
- 否则, 我们将计算max_so_far和(sum – min_so_far)的最大值并将其返回。
下面给出上述方法的C++实现。
C ++
//C++ program for maximum contiguous circular sum problem
#include <bits/stdc++.h>
using namespace std;
//The function returns maximum
//circular contiguous sum in a[]
int maxCircularSum( int a[], int n)
{
//Corner Case
if (n == 1)
return a[0];
//Initialize sum variable which store total sum of the array.
int sum = 0;
for ( int i = 0; i <n; i++) {
sum += a[i];
}
//Initialize every variable with first value of array.
int curr_max = a[0], max_so_far = a[0], curr_min = a[0], min_so_far = a[0];
//Concept of Kadane's Algorithm
for ( int i = 1; i <n; i++) {
//Kadane's Algorithm to find Maximum subarray sum.
curr_max = max(curr_max + a[i], a[i]);
max_so_far = max(max_so_far, curr_max);
//Kadane's Algorithm to find Minimum subarray sum.
curr_min = min(curr_min + a[i], a[i]);
min_so_far = min(min_so_far, curr_min);
}
if (min_so_far == sum)
return max_so_far;
//returning the maximum value
return max(max_so_far, sum - min_so_far);
}
/* Driver program to test maxCircularSum() */
int main()
{
int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
int n = sizeof (a) /sizeof (a[0]);
cout <<"Maximum circular sum is " <<maxCircularSum(a, n) <<endl;
return 0;
}
输出如下:
Maximum circular sum is 31
复杂度分析:
- 时间复杂度:O(n), 其中n是输入数组中元素的数量。
由于仅需要数组的线性遍历。 - 辅助空间:O(1)。
由于不需要额外的空间。