Remove Zero Sum Consecutive Nodes from Linked List Solutions in Go

package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// ListNode define
type ListNode = structures.ListNode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */

// 解法一
func removeZeroSumSublists(head *ListNode) *ListNode {
	// 计算累加和，和作为 key 存在 map 中，value 存那个节点的指针。如果字典中出现了重复的和，代表出现了和为 0 的段。
	sum, sumMap, cur := 0, make(map[int]*ListNode), head
	// 字典中增加 0 这个特殊值，是为了防止最终链表全部消除完
	sumMap[0] = nil
	for cur != nil {
		sum = sum + cur.Val
		if ptr, ok := sumMap[sum]; ok {
			// 在字典中找到了重复的和，代表 [ptr, tmp] 中间的是和为 0 的段，要删除的就是这一段。
			// 同时删除 map 中中间这一段的和
			if ptr != nil {
				iter := ptr.Next
				tmpSum := sum + iter.Val
				for tmpSum != sum {
					// 删除中间为 0 的那一段，tmpSum 不断的累加删除 map 中的和
					delete(sumMap, tmpSum)
					iter = iter.Next
					tmpSum = tmpSum + iter.Val
				}
				ptr.Next = cur.Next
			} else {
				head = cur.Next
				sumMap = make(map[int]*ListNode)
				sumMap[0] = nil
			}
		} else {
			sumMap[sum] = cur
		}
		cur = cur.Next
	}
	return head
}

// 解法二 暴力解法
func removeZeroSumSublists1(head *ListNode) *ListNode {
	if head == nil {
		return nil
	}
	h, prefixSumMap, sum, counter, lastNode := head, map[int]int{}, 0, 0, &ListNode{Val: 1010}
	for h != nil {
		for h != nil {
			sum += h.Val
			counter++
			if v, ok := prefixSumMap[sum]; ok {
				lastNode, counter = h, v
				break
			}
			if sum == 0 {
				head = h.Next
				break
			}
			prefixSumMap[sum] = counter
			h = h.Next
		}
		if lastNode.Val != 1010 {
			h = head
			for counter > 1 {
				counter--
				h = h.Next
			}
			h.Next = lastNode.Next
		}
		if h == nil {
			break
		} else {
			h, prefixSumMap, sum, counter, lastNode = head, map[int]int{}, 0, 0, &ListNode{Val: 1010}
		}
	}
	return head
}