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Formatted question description: https://leetcode.ca/all/1171.html

1171. Remove Zero Sum Consecutive Nodes from Linked List

Level

Medium

Description

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:

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

Solution

Create a dummyHead and set dummyHead.next = head. Initialize temp = dummyHead. Each time delete zero sum consecutive nodes starting from temp.next. If there are nodes that are deleted, set temp.next to be the next node after deletion. Otherwise, set temp = temp.next.

Finally, return dummyHead.next.

  • /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    class Solution {
        public ListNode removeZeroSumSublists(ListNode head) {
            ListNode dummyHead = new ListNode(0);
            dummyHead.next = head;
            ListNode temp = dummyHead;
            while (temp != null) {
                ListNode zeroLast = removeZero(temp.next);
                if (zeroLast == null)
                    temp = temp.next;
                else
                    temp.next = zeroLast.next;
            }
            return dummyHead.next;
        }
    
        public ListNode removeZero(ListNode node) {
            ListNode temp = node;
            int sum = 0;
            while (temp != null) {
                sum += temp.val;
                if (sum == 0)
                    return temp;
                else
                    temp = temp.next;
            }
            return null;
        }
    }
    
  • // OJ: https://leetcode.com/problems/remove-zero-sum-consecutive-nodes-from-linked-list/
    // Time: O(N)
    // Space: O(N)
    class Solution {
    public:
        ListNode* removeZeroSumSublists(ListNode* head) {
            ListNode dummy;
            dummy.next = head;
            unordered_map<int, ListNode*> m{ {0, &dummy} };
            auto node = head;
            int sum = 0;
            while (node) {
                sum += node->val;
                if (m.count(sum)) {
                    int s = sum;
                    for (auto p = m[sum]->next; p != node; p = p->next) {
                        s += p->val;
                        m.erase(s);
                    }
                    m[sum]->next = node->next;
                if (m.count(sum)) {
                } else {
                    m[sum] = node;
                }
                node = node->next;
            }
            return dummy.next;
        }
    };
    
  • # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, val=0, next=None):
    #         self.val = val
    #         self.next = next
    class Solution:
        def removeZeroSumSublists(self, head: Optional[ListNode]) -> Optional[ListNode]:
            dummy = ListNode(next=head)
            last = {}
            s, cur = 0, dummy
            while cur:
                s += cur.val
                last[s] = cur
                cur = cur.next
            s, cur = 0, dummy
            while cur:
                s += cur.val
                cur.next = last[s].next
                cur = cur.next
            return dummy.next
    
    ############
    
    # 1171. Remove Zero Sum Consecutive Nodes from Linked List
    # https://leetcode.com/problems/remove-zero-sum-consecutive-nodes-from-linked-list/
    
    # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, val=0, next=None):
    #         self.val = val
    #         self.next = next
    class Solution:
        def removeZeroSumSublists(self, head: ListNode) -> ListNode:
            prefix = 0
            seen = {}
            seen[0] = dummy = ListNode(0)
            dummy.next = head
            
            while head:
                prefix += head.val
                seen[prefix] = head
                head = head.next
            
            prefix = 0
            head = dummy
            
            while head:
                prefix += head.val
                head.next = seen[prefix].next
                head = head.next
            
            return dummy.next
    
    
  • /**
     * Definition for singly-linked list.
     * type ListNode struct {
     *     Val int
     *     Next *ListNode
     * }
     */
    func removeZeroSumSublists(head *ListNode) *ListNode {
    	dummy := &ListNode{0, head}
    	last := map[int]*ListNode{}
    	cur := dummy
    	s := 0
    	for cur != nil {
    		s += cur.Val
    		last[s] = cur
    		cur = cur.Next
    	}
    	s = 0
    	cur = dummy
    	for cur != nil {
    		s += cur.Val
    		cur.Next = last[s].Next
    		cur = cur.Next
    	}
    	return dummy.Next
    }
    
  • /**
     * Definition for singly-linked list.
     * class ListNode {
     *     val: number
     *     next: ListNode | null
     *     constructor(val?: number, next?: ListNode | null) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.next = (next===undefined ? null : next)
     *     }
     * }
     */
    
    function removeZeroSumSublists(head: ListNode | null): ListNode | null {
        const dummy = new ListNode(0, head);
        const last = new Map<number, ListNode>();
        let s = 0;
        for (let cur = dummy; cur; cur = cur.next) {
            s += cur.val;
            last.set(s, cur);
        }
        s = 0;
        for (let cur = dummy; cur; cur = cur.next) {
            s += cur.val;
            cur.next = last.get(s)!.next;
        }
        return dummy.next;
    }
    
    

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