<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* removeZeroSumSublists(ListNode* head) {
        if(head==NULL){
            return head;
        }
        ListNode* temp = head;
        vector<int> nums;
        vector<int> finalList;
        while(temp!=NULL){
            nums.push_back(temp->val);
            temp = temp->next;
        }

        int n = nums.size();
        int sum = 0;
        for(int i=0; i<n; i++){
            sum = nums[i];
            if(sum==0){
                continue;  
            }
            int start = i;
            for(int j=i+1; j<n; j++){
                sum += nums[j];
                if(sum==0){
                    i=j;
                    break;
                }
            } 
            if(sum!=0){
                finalList.push_back(nums[start]);
            }      
        }
        if(finalList.size()==0){
            return NULL;
        }

        //convert vector into linked list
        ListNode* ans = new ListNode(0);
        ListNode* curr = ans;
        for(int i=0; i<finalList.size(); i++){
            ListNode* temp = new ListNode(finalList[i]);
            curr->next = temp;
            curr = curr->next;
        }
        return ans->next;
    }
};
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