README_EN.md

1171. Remove Zero Sum Consecutive Nodes from Linked List

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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.

Solutions

Python3

# 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

Java

/**
 * 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) {
        ListNode dummy = new ListNode(0, head);
        Map<Integer, ListNode> last = new HashMap<>();
        int s = 0;
        ListNode cur = dummy;
        while (cur != null) {
            s += cur.val;
            last.put(s, cur);
            cur = cur.next;
        }
        s = 0;
        cur = dummy;
        while (cur != null) {
            s += cur.val;
            cur.next = last.get(s).next;
            cur = cur.next;
        }
        return dummy.next;
    }
}

C++

/**
 * 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) {
        ListNode* dummy = new ListNode(0, head);
        unordered_map<int, ListNode*> last;
        ListNode* cur = dummy;
        int s = 0;
        while (cur) {
            s += cur->val;
            last[s] = cur;
            cur = cur->next;
        }
        s = 0;
        cur = dummy;
        while (cur) {
            s += cur->val;
            cur->next = last[s]->next;
            cur = cur->next;
        }
        return dummy->next;
    }
};

Go

/**
 * 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
}

TypeScript

/**
 * 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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