查找是否存在总和为 0 的子数组

给定一个由正数和负数组成的数组，查找是否存在一个总和为 0 的子数组（大小至少为一个）。

例子 ：

Input: {4, 2, -3, 1, 6}
Output: true 
Explanation:
There is a subarray with zero sum from index 1 to 3.

Input: {4, 2, 0, 1, 6}
Output: true 
Explanation:
There is a subarray with zero sum from index 2 to 2.

Input: {-3, 2, 3, 1, 6}
Output: false

一个简单的解决方案是一个一个地考虑所有子数组并检查每个子数组的总和。我们可以运行两个循环：外循环选择一个起点 i，内循环尝试从 i 开始的所有子数组（见实现）。该方法的时间复杂度为 O(n ² )。
我们也可以使用散列。这个想法是遍历数组，对于每个元素 arr[i]，计算从 0 到 i 的元素的总和（这可以简单地用 sum += arr[i] 来完成）。如果之前已经看到当前和，则存在零和数组。散列用于存储和值，以便我们可以快速存储和并找出之前是否见过当前和。
例子 ：

arr[] = {1, 4, -2, -2, 5, -4, 3}

If we consider all prefix sums, we can
notice that there is a subarray with 0
sum when :
1) Either a prefix sum repeats or
2) Or prefix sum becomes 0.

Prefix sums for above array are:
1, 5, 3, 1, 6, 2, 5

Since prefix sum 1 repeats, we have a subarray
with 0 sum. 

以下是上述方法的实现。

C++

// A C++ program to find if
// there is a zero sum subarray
#include 
using namespace std;
 
bool subArrayExists(int arr[], int n)
{
    unordered_set sumSet;
 
    // Traverse through array
    // and store prefix sums
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum += arr[i];
 
        // If prefix sum is 0 or
        // it is already present
        if (sum == 0
            || sumSet.find(sum)
            != sumSet.end())
            return true;
 
        sumSet.insert(sum);
    }
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { -3, 2, 3, 1, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (subArrayExists(arr, n))
        cout << "Found a subarray with 0 sum";
    else
        cout << "No Such Sub Array Exists!";
    return 0;
}

Java

// A Java program to find
// if there is a zero sum subarray
import java.util.HashSet;
import java.util.Set;
 
class ZeroSumSubarray
{
    // Returns true if arr[]
    // has a subarray with sero sum
    static Boolean subArrayExists(int arr[])
    {
        // Creates an empty hashset hs
        Set hs = new HashSet();
 
        // Initialize sum of elements
        int sum = 0;
 
        // Traverse through the given array
        for (int i = 0; i < arr.length; i++)
        {
            // Add current element to sum
            sum += arr[i];
 
            // Return true in following cases
            // a) Current element is 0
            // b) sum of elements from 0 to i is 0
            // c) sum is already present in hash map
            if (arr[i] == 0
                || sum == 0
                || hs.contains(sum))
                return true;
 
            // Add sum to hash set
            hs.add(sum);
        }
 
        // We reach here only when there is
        // no subarray with 0 sum
        return false;
    }
 
    // Driver code
    public static void main(String arg[])
    {
        int arr[] = { -3, 2, 3, 1, 6 };
        if (subArrayExists(arr))
            System.out.println(
                "Found a subarray with 0 sum");
        else
            System.out.println("No Such Sub Array Exists!");
    }
}

Python3

# A python program to find if
# there is a zero sum subarray
 
 
def subArrayExists(arr, n):
    # traverse through array
    # and store prefix sums
    n_sum = 0
    s = set()
 
    for i in range(n):
        n_sum += arr[i]
 
        # If prefix sum is 0 or
        # it is already present
        if n_sum == 0 or n_sum in s:
            return True
        s.add(n_sum)
 
    return False
 
 
# Driver code
arr = [-3, 2, 3, 1, 6]
n = len(arr)
if subArrayExists(arr, n) == True:
    print("Found a sunbarray with 0 sum")
else:
    print("No Such sub array exits!")
 
# This code is contributed by Shrikant13

C#

// A C# program to find if there
// is a zero sum subarray
using System;
using System.Collections.Generic;
 
class GFG {
    // Returns true if arr[] has
    // a subarray with sero sum
    static bool SubArrayExists(int[] arr)
    {
        // Creates an empty HashSet hM
        HashSet hs = new HashSet();
        // Initialize sum of elements
        int sum = 0;
 
        // Traverse through the given array
        for (int i = 0; i < arr.Length; i++)
        {
            // Add current element to sum
            sum += arr[i];
 
            // Return true in following cases
            // a) Current element is 0
            // b) sum of elements from 0 to i is 0
            // c) sum is already present in hash set
            if (arr[i] == 0
                || sum == 0
                || hs.Contains(sum))
                return true;
 
            // Add sum to hash set
            hs.Add(sum);
        }
 
        // We reach here only when there is
        // no subarray with 0 sum
        return false;
    }
 
    // Main Method
    public static void Main()
    {
        int[] arr = { -3, 2, 3, 1, 6 };
        if (SubArrayExists(arr))
            Console.WriteLine(
                "Found a subarray with 0 sum");
        else
            Console.WriteLine("No Such Sub Array Exists!");
    }
}

Javascript

// A Javascript program to
//  find if there is a zero sum subarray
 
const subArrayExists = (arr) => {
    const sumSet = new Set();
 
    // Traverse through array
    // and store prefix sums
    let sum = 0;
    for (let i = 0 ; i < arr.length ; i++)
    {
        sum += arr[i];
 
        // If prefix sum is 0
        // or it is already present
        if (sum === 0 || sumSet.has(sum))
            return true;
 
        sumSet.add(sum);
    }
    return false;
}
 
// Driver code
 
const arr =  [-3, 2, 3, 1, 6];
if (subArrayExists(arr))
    console.log("Found a subarray with 0 sum");
else
    console.log("No Such Sub Array Exists!");

输出

No Such Sub Array Exists!

该解决方案的时间复杂度可以被认为是 O(n)，前提是我们有良好的散列函数，允许在 O(1) 时间内进行插入和检索操作。
空间复杂度：O(n)。这里我们需要额外的空间让 unordered_set 插入数组元素。

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