1171. Remove Zero Sum Consecutive Nodes from Linked List 

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class Solution {
    public ListNode removeZeroSumSublists(ListNode head) {
        ListNode dummyHead = new ListNode(-1, head);
        Map<Integer, ListNode> prefixSumToNodeMap = new HashMap<>();
        prefixSumToNodeMap.put(0, dummyHead);

        int prefixSum = 0; // 截止至当前元素的前缀和（包含当前元素）

        ListNode curr = head;
        while (curr != null) {
            prefixSum += curr.val;
            ListNode prevNode = prefixSumToNodeMap.get(prefixSum);
            if (prevNode != null) {
                ListNode node = prevNode.next;
                int tempSum = prefixSum;
                while (node.next != curr.next) {
                    tempSum += node.val;
                    prefixSumToNodeMap.remove(tempSum);
                    node = node.next;
                }

                prevNode.next = curr.next;
            } else {
                prefixSumToNodeMap.put(prefixSum, curr);
            }

            curr = curr.next;
        }

        return dummyHead.next;
    }
}
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class Solution {
    public ListNode removeZeroSumSublists(ListNode head) {
        ListNode dummyHead = new ListNode(-1, head);
        Map<Integer, ListNode> prefixSumToNodeMap = new HashMap<>();
        int prefixSum = 0; // 截止至当前元素的前缀和（包含当前元素）
        for (ListNode curr = dummyHead; curr != null; curr = curr.next) {
            prefixSum += curr.val;
            prefixSumToNodeMap.put(prefixSum, curr); // 相同前缀和的保留最后一个出现的节点
        }

        prefixSum = 0;
        for (ListNode curr = dummyHead; curr != null; curr = curr.next) {
            prefixSum += curr.val;
            curr.next = prefixSumToNodeMap.get(prefixSum).next;
        }

        return dummyHead.next;
    }
}

References

1171. Remove Zero Sum Consecutive Nodes from Linked List