[LeetCode 1171] Remove Zero Sum Consecutive Nodes from Linked List

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 1. Two pointers, O(N^2) runtime, O(1) space

1. Have a left and right pointer l and r; Also init a pointer called prevL that is the previous node of l.

2. start l with the first node, move r one node at a time. For each node that r is visiting, there will be the following 2 cases.

a. if the running sum is 0, delete all nodes from l to r inclusively by setting the next pointer of prevL to r‘s next node. Then update l to r.

b. else, update prevL to l.

3. move l to its next node and repeat step 2 until l reaches the end of the linked list.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeZeroSumSublists(ListNode head) {
        ListNode dummy = new ListNode(1000 * 1000 + 1);
        dummy.next = head;
        ListNode prevL = dummy, l = head, r = head;

        while(l != null) {
            r = l;
            int sum = 0;
            boolean skip = false;
            while(r != null) {
                sum += r.val;
                if(sum == 0) {
                    prevL.next = r.next;
                    l = r;
                    skip = true;
                    break;
                }
                r = r.next;
            }
            if(!skip) {
                prevL = l;
            }
            l = l.next;
        }
        return dummy.next;
    }
}

Solution 2. Prefix Sum with HashMap, O(N) runtime and space

1. Create a dummy node of value 0 and a hashmap that maps prefixsum to node.

2. Starting from the dummy node, compute prefix sum that ends at the current node.

If there already exists a prefix sum of the same value in the hashmap, we know all nodes from map.get(prefixSum).next to the current node(inclusive) sum to 0. Delete all these nodes by map.get(prefixSum).next = curr.next.  Since these nodes are deleted, we need to remove all the prefix sums that end at these deleted nodes from the map.

Else, put the current prefix sum and node to the map for later checks.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeZeroSumSublists(ListNode head) {
        ListNode dummy = new ListNode(0), curr = dummy;
        Map<Integer, ListNode> map = new HashMap<>();
        dummy.next = head;
        int prefixSum = 0;
        while(curr != null) {
            prefixSum += curr.val;
            if(map.containsKey(prefixSum)) {
                ListNode node = map.get(prefixSum).next;
                int sum = prefixSum;
                while(node != curr) {
                    sum += node.val;
                    map.remove(sum);
                    node = node.next;
                }
                map.get(prefixSum).next = curr.next;
            }
            else {
                map.put(prefixSum, curr);
            }
            curr = curr.next;
        }
        return dummy.next;
    }
}

原文地址：https://www.cnblogs.com/lz87/p/11605711.html