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: [1]
Constraints:
- The given linked list will contain between
1and1000nodes. - Each node in the linked list has
-1000 <= node.val <= 1000.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public ListNode removeZeroSumSublists(ListNode head) {
ListNode dummy = new ListNode(0);
ListNode curr = head;
dummy.next = head;
Map < Integer, ListNode> 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;
}
}
Input
head = [1,2,-3,3,1]
Output
[3,1]
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const removeZeroSumSublists = function(head) {
let dummy = new ListNode(0), cur = dummy;
dummy.next = head;
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;
};
Input
head = [1,2,-3,3,1]
Output
[3,1]