如何删除链表中相加为0的连续元素

Question

我正在编写一个 Python 代码来删除链表中的那些连续元素，这些元素加起来为 0

链表定义如下：

class Node:
    def __init__(self, val, next=None):
        self.value = val
        self.next = next

node = Node(10)
node.next = Node(5)
node.next.next = Node(-3)
node.next.next.next = Node(-3)
node.next.next.next.next = Node(1)
node.next.next.next.next.next = Node(4)
node.next.next.next.next.next.next = Node(-4)

从以上数据中，5 -> -3 -> -3 -> 1和4 -> -4加起来是0，需要剔除。

遍历元素后，如

def removeConsecutiveSumTo0(node):
    start = node
    while start:
        mod = False
        total = 0
        end = start

        while end:
            total += end.value
            if total == 0:
                start = end
                mod = True
                break
            end = end.next

        if mod == False:
            res = start

        start = start.next

    return res

node = removeConsecutiveSumTo0(node)
while node:
    print (node.value, end=' ')
    node = node.next
# 10 (Expected output)

我无法创建包含加起来为 0 的连续元素的子集。因为它是 NP-Complete problem 所讨论的 and here。我如何设计算法来找到解决方案？

Answer 1

您可以尝试递归或嵌套循环，因为您应该尝试从每个节点开始计算总和。一个天真的实现可能如下：

def removeConsecutiveSumTo0(node):
  start = node
  while start.next:
    total = 0
    cur = start.next
    while cur:
      total += cur.value
      if total == 0:
        start.next = cur.next
        break
      cur = cur.next
    else:
      start = start.next

如何删除链表中相加为0的连续元素

How to delete consecutive elements in a linked list which add up to 0

python

algorithm

linked-list

subset-sum

np-complete