<p>Given the <code>head</code> of a linked list, we repeatedly delete consecutive sequences of nodes that sum to <code>0</code> until there are no such sequences.</p>

<p>After doing so, return the head of the final linked list.&nbsp; You may return any such answer.</p>

<p>&nbsp;</p>
<p>(Note that in the examples below, all sequences are serializations of <code>ListNode</code> objects.)</p>

<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,-3,3,1]
<strong>Output:</strong> [3,1]
<strong>Note:</strong> The answer [1,2,1] would also be accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,3,-3,4]
<strong>Output:</strong> [1,2,4]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,3,-3,-2]
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The given linked list will contain between <code>1</code> and <code>1000</code> nodes.</li>
	<li>Each node in the linked list has <code>-1000 &lt;= node.val &lt;= 1000</code>.</li>
</ul>

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* convertToLL(vector<int>v)
    {
        if(v.empty())
        return NULL;
        ListNode* head=new ListNode(v[0]),*temp=head;
        for(int i=1;i<v.size();i++)
        {
            ListNode* newnode=new ListNode(v[i]);
            temp->next=newnode;
            temp=newnode;
        }
        return head;
    }
public:
    ListNode* removeZeroSumSublists(ListNode* head) {
        vector<int>v;
        ListNode*temp=head;
        while(temp)
        {
            v.push_back(temp->val);
            temp=temp->next;
        }
        for(int i=0;i<v.size();i++)
        {
            int sum=0;
            for(int j=i;j<v.size();j++)
            {
                sum+=v[j];
                if(sum==0)
                {
                    v.erase(v.begin()+i,v.begin()+j+1);
                    i--;
                    break;
                }
            }
        }
        return convertToLL(v);   
    }
};
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