## Algorithm

Problem Name: 1171. Remove Zero Sum Consecutive Nodes from Linked List

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

After doing so, return the head of the final linked list.  You may return any such answer.

(Note that in the examples below, all sequences are serializations of `ListNode` objects.)

Example 1:

```Input: head = [1,2,-3,3,1]
Output: [3,1]
Note: The answer [1,2,1] would also be accepted.
```

Example 2:

```Input: head = [1,2,3,-3,4]
Output: [1,2,4]
```

Example 3:

```Input: head = [1,2,3,-3,-2]
Output: 
```

Constraints:

• The given linked list will contain between `1` and `1000` nodes.
• Each node in the linked list has `-1000 <= node.val <= 1000`.

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
ListNode dummy = new ListNode(0);
Map map = new HashMap<>();
int prefixSum = 0;
map.put(0, dummy);
while (curr != null) {
prefixSum += curr.val;
map.put(prefixSum, curr);
curr = curr.next;
}
prefixSum = 0;
curr = dummy;
while (curr != null) {
prefixSum += curr.val;
curr.next = map.get(prefixSum).next;
curr = curr.next;
}
return dummy.next;
}
}
``````
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Input

cmd

Output

cmd
[3,1]

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
let dummy = new ListNode(0), cur = dummy;
let prefix = 0;
let m = new Map();
while (cur != null) {
prefix += cur.val;
if (m.has(prefix)) {
cur =  m.get(prefix).next;
let p = prefix + cur.val;
while (p != prefix) {
m.delete(p);
cur = cur.next;
p += cur.val;
}
m.get(prefix).next = cur.next;
} else {
m.set(prefix, cur);
}
cur = cur.next;
}
return dummy.next;
};
``````
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Input

cmd