# 1171. Remove Zero Sum Consecutive Nodes from Linked List 从链表中删去总和值为零的连续节点

@TOC

## # 题目描述

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:

1. The given linked list will contain between `1` and `1000` nodes.
2. Each node in the linked list has `-1000 <= node.val <= 1000`.

## # 解题方法

### # preSum + 字典

C++代码如下：

``````/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeZeroSumSublists(ListNode* head) {
unordered_map<int, ListNode*> m;
ListNode* dummy = new ListNode(-10000);
int preSum = 0;
m[0] = dummy;
ListNode* cur = head;
while (cur) {
preSum += cur->val;
if (m.count(preSum)) {
ListNode* pre = m[preSum];
ListNode* cur = pre->next;
int p = preSum + cur->val;
while (p != preSum) {
m.erase(p);
cur = cur->next;
p += cur->val;
}
pre->next = cur->next;
} else {
m[preSum] = cur;
}
cur = cur->next;
}
return dummy->next;
}
};
``````

## # 日期

2019 年 9 月 27 日 —— 昨天面快手，竟然是纯刷题