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: [1]

 

Constraints:


Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeZeroSumSublists(ListNode head) {
        /* 
            Iterate for the first time,
            calculate the prefix sum,
            and save the it to seen[prefix]

            Iterate for the second time,
            calculate the prefix sum,
            and directly skip to last occurrence of this prefix
        */
        int prefix = 0;
        Map<Integer, ListNode> map = new HashMap();
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        for (ListNode i = dummy; i != null; i = i.next) {
            prefix += i.val;
            map.put(prefix, i);
        }
        prefix = 0;
        for (ListNode i = dummy; i != null; i = i.next) {
            prefix += i.val;
            i.next = map.get(prefix).next;
        }
        return dummy.next;
    }
}