Add solution #1171

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,164 LeetCode solutions in JavaScript
+# 1,165 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -912,6 +912,7 @@
 1163|[Last Substring in Lexicographical Order](./solutions/1163-last-substring-in-lexicographical-order.js)|Hard|
 1169|[Invalid Transactions](./solutions/1169-invalid-transactions.js)|Medium|
 1170|[Compare Strings by Frequency of the Smallest Character](./solutions/1170-compare-strings-by-frequency-of-the-smallest-character.js)|Medium|
+1171|[Remove Zero Sum Consecutive Nodes from Linked List](./solutions/1171-remove-zero-sum-consecutive-nodes-from-linked-list.js)|Medium|
 1189|[Maximum Number of Balloons](./solutions/1189-maximum-number-of-balloons.js)|Easy|
 1200|[Minimum Absolute Difference](./solutions/1200-minimum-absolute-difference.js)|Easy|
 1206|[Design Skiplist](./solutions/1206-design-skiplist.js)|Hard|

solutions/1171-remove-zero-sum-consecutive-nodes-from-linked-list.js

@@ -0,0 +1,48 @@
+/**
+ * 1171. Remove Zero Sum Consecutive Nodes from Linked List
+ * https://leetcode.com/problems/remove-zero-sum-consecutive-nodes-from-linked-list/
+ * Difficulty: Medium
+ *
+ * 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.)
+ */
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var removeZeroSumSublists = function(head) {
+  const node = new ListNode(0, head);
+  const map = new Map();
+  let prefixSum = 0;
+
+  let current = node;
+  while (current) {
+    prefixSum += current.val;
+    map.set(prefixSum, current);
+    current = current.next;
+  }
+
+  prefixSum = 0;
+  current = node;
+  while (current) {
+    prefixSum += current.val;
+    if (map.has(prefixSum) && map.get(prefixSum) !== current) {
+      current.next = map.get(prefixSum).next;
+    }
+    current = current.next;
+  }
+
+  return node.next;
+};