Solution for 3Sum #241

Refactor comments in 3Sum solution for clarity and consistency

Author: dohyeon2

3sum/dohyeon2.java

@@ -0,0 +1,57 @@
+import java.util.Arrays;
+import java.util.ArrayList;
+import java.util.List;
+
+class Solution {
+    // TC: O(n^2)
+    // SC: O(1)
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> answer = new ArrayList<List<Integer>>();
+
+        // Sort the array to use two-pointer technique.
+        Arrays.sort(nums);
+
+        // Exclude the last two elements from the loop
+        // since the two pointers are involved in the iteration.
+        for (int i = 0; i < nums.length - 2; i++) {
+            if (i - 1 >= 0 && nums[i] == nums[i - 1]) {
+                // Skip if this number is the same as the previous one,
+                // so that we can avoid duplicate triplets.
+                continue;
+            }
+
+            int left = i + 1;
+            int right = nums.length - 1;
+
+            while (left < right) {
+                int sum = nums[i] + nums[left] + nums[right];
+                if (sum == 0) {
+                    ArrayList<Integer> list = new ArrayList<>(
+                            Arrays.asList(nums[i], nums[left], nums[right]));
+                    answer.add(list);
+
+                    // According to the problem, we need to avoid duplicate triplets.
+                    // Therefore, this loop is needed.
+                    while (left < right && nums[left] == nums[left + 1]) {
+                        left++;
+                    }
+                    while (left < right && nums[right] == nums[right - 1]) {
+                        right--;
+                    }
+
+                    left++;
+                    right--;
+                }
+                if (sum < 0) {
+                    left++;
+                }
+                if (sum > 0) {
+                    right--;
+                }
+            }
+
+        }
+
+        return answer;
+    }
+}