S
S
Study Notes
Search
K

1171.remove-zero-sum-consecutive-nodes-from-linked-list-en

Problem

Problem 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

This problem is typical prefixSum problem, if prefixSum seen, then all elements between two same prefixSum sum = 0.
Here, we can use HashMapkey as prefixSum, value as current Node, if prefixSum already seen in HashMap, then delete all nodes between two same prefixSum. and also delete relative prefixSum in HashMap. (We already removed nodes from head, so that remove cumulative sum with those nodes. otherwise, it is incorrect.)
Here, we use dummy node, ListNode(0), (as shown in below pic), as initiating HashMap, put(0, ListNode(0)) into HashMap, iterate ListNode
  • if prefixSum in HashMap, delete all nodes
  • if prefixSum not in HashMap, put prefixSum and currNode into HashMap. map.put(prefixSum, currNode).
举例:head = [1, 2, 3, -2, -3, 9]
1171 example remove zero

Complexity Analysis

  • Time Complexity: O(n) - n is number of ListNode
  • Space Complexity: O(n) - HashMap

Key Points

  • calculate prefixSum
  • use HashMap keep track all prefixSum and currNode.
  • Initialize HashMap,put0 (here usedummy node
  • Delete all Nodes between two same prefixSum, delete related prefixSum from HashMap.
  • return dummy.next;.

Code (Java/Python3)

Java Code
class RemoveZeroInLinkedList {
public static ListNode removeZeroSumSublists(ListNode head) {
ListNode dummy = new ListNode(0);
ListNode currNode = dummy;
dummy.next = head;
int prefixSum = 0;
Map<Integer, ListNode> prefixSumMap = new HashMap<>();
while (currNode != null) {
prefixSum += currNode.val;
// seen prefixSum in HashMap, remove all nodes and relative prefixSum from HashMap
if (prefixSumMap.containsKey(prefixSum)) {
currNode = prefixSumMap.get(prefixSum).next;
int sum = prefixSum + currNode.val;
while (sum != prefixSum) {
prefixSumMap.remove(sum);
currNode = currNode.next;
sum += currNode.val;
}
prefixSumMap.get(prefixSum).next = currNode.next;
} else {
// prefixSum not seen in HashMap, add into HashMap
prefixSumMap.put(prefixSum, currNode);
}
currNode = currNode.next;
}
return dummy.next;
}
}
Python3 code
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeZeroSumSublists(self, head: ListNode) -> ListNode:
dummy = curr_node = ListNode(0)
dummy.next = head
prefix_sum = 0
prefix_sum_map = {}
while curr_node:
prefix_sum += curr_node.val
if prefix_sum in prefix_sum_map:
curr_node = prefix_sum_map.get(prefix_sum).next
sum = prefix_sum + curr_node.val
while sum != prefix_sum and sum in prefix_sum_map:
del prefix_sum_map[sum]
curr_node = curr_node.next
sum += curr_node.val
prefix_sum_map[prefix_sum].next = curr_node.next
else:
prefix_sum_map[prefix_sum] = curr_node
curr_node = curr_node.next
return dummy.next