[LeetCode1171] Remove Zero Sum Consecutive Nodes from Linked List

题目链接

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

if prefixSum[i] = prefixSum[j], erase from i to j.

Code

One Hash (WA)

One Hash is not Enough
with the example
[1,3,2,-3,-2,5,5,-5,1]

# 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:
        dummyNode = ListNode(0)
        dummyNode.next = currentNode = head
        prefixSum = 0
        prefixMap = {0:dummyNode}
        while currentNode:
            prefixSum += currentNode.val
            if prefixSum in prefixMap:
                prefixMap[prefixSum].next = currentNode.next
            else:
                prefixMap[prefixSum] = currentNode
            currentNode = currentNode.next
        return dummyNode.next

One Hash with rollback

# 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:
        dummyNode = ListNode(0)
        dummyNode.next, currentNode,prefixSum, prefixMap = head, head, 0, {0:dummyNode}
        while currentNode:
            prefixSum += currentNode.val
            if prefixSum in prefixMap:
                prefixMap[prefixSum].next = currentNode.next
                currentNode, prefixSum = dummyNode.next, 0
                prefixMap = {0:dummyNode}
            else:
                prefixMap[prefixSum] = currentNode
                currentNode = currentNode.next
        return dummyNode.next

One Hash with loop

# 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:
        dummyNode = ListNode(0)
        dummyNode.next = head
        currentNode, prefixSum, prefixMap = dummyNode, 0, {}
        while currentNode:
            prefixSum += currentNode.val
            prefixMap[prefixSum] = currentNode
            currentNode = currentNode.next
        currentNode, prefixSum = dummyNode, 0
        while currentNode:
            prefixSum += currentNode.val
            if prefixSum in prefixMap:
                currentNode.next = prefixMap[prefixSum].next
            currentNode = currentNode.next
        return dummyNode.next
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