给定一个由N个正数组成的数组，任务是要找到一个连续的子数组(LR)，使得a [L] = a [R]和a [L] + a [L + 1] +…+ a [R ]为最大。
例子：
Input: arr[] = {1, 3, 2, 2, 3}
Output: 10
Subarray [3, 2, 2, 3] starts and ends with 3 and has sum = 10
Input: arr[] = {1, 3, 2, 2, 3}
Output: 10
方法：对于数组中的每个元素，让我们找到2个值：数组中的第一个(最左)和数组中的最后一个(最右)。由于所有数字均为正数，因此增加项数只会增加总和。因此，对于数组中的每个数字，我们都找到它最左边和最右边出现之间的总和，可以使用前缀总和快速完成。我们可以跟踪到目前为止找到的最大值并最终打印出来。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to find the maximum sum
int maxValue(int a[], int n)
{
unordered_map first, last;
int pr[n];
pr[0] = a[0];
for (int i = 1; i < n; i++) {
// Build prefix sum array
pr[i] = pr[i - 1] + a[i];
// If the value hasn't been encountered before,
// It is the first occurrence
if (first[a[i]] == 0)
first[a[i]] = i;
// Keep updating the last occurrence
last[a[i]] = i;
}
int ans = 0;
// Find the maximum sum with same first and last value
for (int i = 0; i < n; i++) {
int start = first[a[i]];
int end = last[a[i]];
ans = max(ans, pr[end] - pr[start - 1]);
}
return ans;
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 5, 2, 4, 18, 2, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxValue(arr, n);
return 0;
} Java
// Java implementation of the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the maximum sum
static int maxValue(int a[], int n)
{
HashMap first = new HashMap<>();
HashMap last = new HashMap<>();
for (int i = 0; i < n; i++) {
first.put(a[i], 0);
last.put(a[i], 0);
}
int[] pr = new int[n];
pr[0] = a[0];
for (int i = 1; i < n; i++) {
// Build prefix sum array
pr[i] = pr[i - 1] + a[i];
// If the value hasn't been encountered before,
// It is the first occurrence
if (Integer.parseInt(String.valueOf(first.get(a[i]))) == 0)
first.put(a[i], i);
// Keep updating the last occurrence
last.put(a[i], i);
}
int ans = 0;
// Find the maximum sum with same first and last value
for (int i = 0; i < n; i++) {
int start = Integer.parseInt(String.valueOf(first.get(a[i])));
int end = Integer.parseInt(String.valueOf(last.get(a[i])));
if (start != 0)
ans = Math.max(ans, pr[end] - pr[start - 1]);
}
return ans;
}
// Driver Code
public static void main(String args[])
{
int[] arr = { 1, 3, 5, 2, 4, 18, 2, 3 };
int n = arr.length;
System.out.print(maxValue(arr, n));
}
}
// This code is contributed by rachana soma Python3
# Python3 implementation of the above approach
from collections import defaultdict
# Function to find the maximum sum
def maxValue(a, n):
first = defaultdict(lambda:0)
last = defaultdict(lambda:0)
pr = [None] * n
pr[0] = a[0]
for i in range(1, n):
# Build prefix sum array
pr[i] = pr[i - 1] + a[i]
# If the value hasn't been encountered before,
# It is the first occurrence
if first[a[i]] == 0:
first[a[i]] = i
# Keep updating the last occurrence
last[a[i]] = i
ans = 0
# Find the maximum sum with same first and last value
for i in range(0, n):
start = first[a[i]]
end = last[a[i]]
ans = max(ans, pr[end] - pr[start - 1])
return ans
# Driver Code
if __name__ == "__main__":
arr = [1, 3, 5, 2, 4, 18, 2, 3]
n = len(arr)
print(maxValue(arr, n))
# This code is contributed by Rituraj JainC#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the maximum sum
static int maxValue(int []a, int n)
{
Dictionary first = new Dictionary();
Dictionary last = new Dictionary();
for (int i = 0; i < n; i++)
{
first[a[i]] = 0;
last[a[i]] = 0;
}
int[] pr = new int[n];
pr[0] = a[0];
for (int i = 1; i < n; i++)
{
// Build prefix sum array
pr[i] = pr[i - 1] + a[i];
// If the value hasn't been encountered before,
// It is the first occurrence
if (first[a[i]] == 0)
first[a[i]] = i;
// Keep updating the last occurrence
last[a[i]] = i;
}
int ans = 0;
// Find the maximum sum with
// same first and last value
for (int i = 0; i < n; i++)
{
int start = first[a[i]];
int end = last[a[i]];
if (start != 0)
ans = Math.Max(ans, pr[end] - pr[start - 1]);
}
return ans;
}
// Driver Code
static void Main()
{
int[] arr = { 1, 3, 5, 2, 4, 18, 2, 3 };
int n = arr.Length;
Console.Write(maxValue(arr, n));
}
}
// This code is contributed by mohit kumar 输出：
37
时间复杂度： O(N)