Add solution #15

Author: JoshCrozier

README.md

@@ -20,6 +20,7 @@
 12|[Integer to Roman](./0012-integer-to-roman.js)|Medium|
 13|[Roman to Integer](./0013-roman-to-integer.js)|Easy|
 14|[Longest Common Prefix](./0014-longest-common-prefix.js)|Easy|
+15|[3Sum](./0015-3sum.js)|Medium|
 17|[Letter Combinations of a Phone Number](./0017-letter-combinations-of-a-phone-number.js)|Medium|
 19|[Remove Nth Node From End of List](./0019-remove-nth-node-from-end-of-list.js)|Medium|
 20|[Valid Parentheses](./0020-valid-parentheses.js)|Easy|

solutions/0015-3sum.js

@@ -0,0 +1,38 @@
+/**
+ * 15. 3Sum
+ * https://leetcode.com/problems/3sum/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j,
+ * i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
+ *
+ * Notice that the solution set must not contain duplicate triplets.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var threeSum = function(nums) {
+  const result = new Set();
+
+  nums.sort((a, b) => a - b);
+
+  for (let i = 0; i < nums.length - 2; i++) {
+    let j = i + 1;
+    let k = nums.length - 1;
+
+    while (j < k) {
+      const sum = nums[i] + nums[j] + nums[k];
+      if (!sum) {
+        result.add(JSON.stringify([nums[i], nums[j++], nums[k--]]));
+      } else if (sum > 0) {
+        k--;
+      } else {
+        j++;
+      }
+    }
+  }
+
+  return [...result].map(JSON.parse);
+};