给定一个整数数组arr[] ，任务是找到非负元素的最大总和连续子数组并返回其总和。
例子：
Input: arr[] = {1, 4, -3, 9, 5, -6}
Output: 14
Explanation:
Subarray [9, 5] is the subarray having maximum sum with all non-negative elements.
Input: arr[] = {12, 0, 10, 3, 11}
Output: 36
天真的方法：
最简单的方法是生成所有只有非负元素的子数组，同时遍历子数组并计算每个有效子数组的总和并更新最大和。
时间复杂度： O(N^2)
有效的方法：
为了优化上述方法，遍历数组，对于遇到的每个非负元素，继续计算总和。对于遇到的每个负元素，在与当前总和比较后更新最大总和。将总和重置为 0 并继续下一个元素。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to return Largest Sum Contiguous
// Subarray having non-negative number
int maxNonNegativeSubArray(int A[], int N)
{
// Length of given array
int l = N;
int sum = 0, i = 0;
int Max = -1;
// Traversing array
while (i < l)
{
// Increment i counter to avoid
// negative elements
while (i < l && A[i] < 0)
{
i++;
continue;
}
// Calculating sum of contiguous
// subarray of non-negative
// elements
while (i < l && 0 <= A[i])
{
sum += A[i++];
// Update the maximum sum
Max = max(Max, sum);
}
// Reset sum
sum = 0;
}
// Return the maximum sum
return Max;
}
// Driver code
int main()
{
int arr[] = { 1, 4, -3, 9, 5, -6 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << maxNonNegativeSubArray(arr, N);
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Function to return Largest Sum Contiguous
// Subarray having non-negative number
static int maxNonNegativeSubArray(int[] A)
{
// Length of given array
int l = A.length;
int sum = 0, i = 0;
int max = -1;
// Traversing array
while (i < l) {
// Increment i counter to avoid
// negative elements
while (i < l && A[i] < 0) {
i++;
continue;
}
// Calculating sum of contiguous
// subarray of non-negative
// elements
while (i < l && 0 <= A[i]) {
sum += A[i++];
// Update the maximum sum
max = Math.max(max, sum);
}
// Reset sum
sum = 0;
}
// Return the maximum sum
return max;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 4, -3, 9, 5, -6 };
System.out.println(maxNonNegativeSubArray(
arr));
}
}Python3
# Python3 program for the above approach
import math
# Function to return Largest Sum Contiguous
# Subarray having non-negative number
def maxNonNegativeSubArray(A, N):
# Length of given array
l = N
sum = 0
i = 0
Max = -1
# Traversing array
while (i < l):
# Increment i counter to avoid
# negative elements
while (i < l and A[i] < 0):
i += 1
continue
# Calculating sum of contiguous
# subarray of non-negative
# elements
while (i < l and 0 <= A[i]):
sum += A[i]
i += 1
# Update the maximum sum
Max = max(Max, sum)
# Reset sum
sum = 0;
# Return the maximum sum
return Max
# Driver code
arr = [ 1, 4, -3, 9, 5, -6 ]
# Length of array
N = len(arr)
print(maxNonNegativeSubArray(arr, N))
# This code is contributed by sanjoy_62C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to return Largest Sum Contiguous
// Subarray having non-negative number
static int maxNonNegativeSubArray(int[] A)
{
// Length of given array
int l = A.Length;
int sum = 0, i = 0;
int max = -1;
// Traversing array
while (i < l)
{
// Increment i counter to avoid
// negative elements
while (i < l && A[i] < 0)
{
i++;
continue;
}
// Calculating sum of contiguous
// subarray of non-negative
// elements
while (i < l && 0 <= A[i])
{
sum += A[i++];
// Update the maximum sum
max = Math.Max(max, sum);
}
// Reset sum
sum = 0;
}
// Return the maximum sum
return max;
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 4, -3, 9, 5, -6 };
Console.Write(maxNonNegativeSubArray(arr));
}
}
// This code is contributed by chitranayalJavascript
C++
#include
using namespace std;
int main()
{
int arr[] = {1, 4, -3, 9, 5, -6};
int n = sizeof(arr)/sizeof(arr[0]);
int max_so_far=0, max_right_here = 0;
int start = 0, end = 0, s=0;
for(int i=0; i max_so_far)
{
max_so_far = max_right_here;
start = s;
end = i;
}
}
cout<<("Sub Array : ");
for(int i = start; i <= end; i++)
{
cout< Java
public class maxNonNegativeSubArray {
public static void main (String[] args) throws java.lang.Exception{
int[] arr = {1, 4, -3, 9, 5, -6};
int max_so_far=0, max_right_here = 0;
int start = 0, end = 0, s=0;
for(int i=0; i max_so_far){
max_so_far = max_right_here;
start = s;
end = i;
}
}
System.out.print("Sub Array : ");
for(int i = start; i <= end; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
System.out.print("Largest Sum : " + max_so_far);
}
} 输出
14时间复杂度：O(N)
辅助空间：O(1)
更简单的方法：
C++
#include
using namespace std;
int main()
{
int arr[] = {1, 4, -3, 9, 5, -6};
int n = sizeof(arr)/sizeof(arr[0]);
int max_so_far=0, max_right_here = 0;
int start = 0, end = 0, s=0;
for(int i=0; i max_so_far)
{
max_so_far = max_right_here;
start = s;
end = i;
}
}
cout<<("Sub Array : ");
for(int i = start; i <= end; i++)
{
cout< Java
public class maxNonNegativeSubArray {
public static void main (String[] args) throws java.lang.Exception{
int[] arr = {1, 4, -3, 9, 5, -6};
int max_so_far=0, max_right_here = 0;
int start = 0, end = 0, s=0;
for(int i=0; i max_so_far){
max_so_far = max_right_here;
start = s;
end = i;
}
}
System.out.print("Sub Array : ");
for(int i = start; i <= end; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
System.out.print("Largest Sum : " + max_so_far);
}
}
输出
Sub Array : 9 5
Largest Sum : 14时间复杂度：O(n)
空间复杂度：O(1)
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