diff --git a/include/s0015_3sum.hpp b/include/s0015_3sum.hpp
new file mode 100644
index 0000000..b1a4607
--- /dev/null
+++ b/include/s0015_3sum.hpp
@@ -0,0 +1,20 @@
+#ifndef S0015_3SUM
+#define S0015_3SUM
+
+#include <vector>
+#include <unordered_map>
+#include <algorithm>
+
+using namespace std;
+
+class Solution1 {
+ public:
+  vector<vector<int>> threeSum(vector<int>& nums);
+};
+
+class Solution2 {
+ public:
+  vector<vector<int>> threeSum(vector<int>& nums);
+};
+
+#endif
diff --git a/src/s0015_3sum.cpp b/src/s0015_3sum.cpp
new file mode 100644
index 0000000..c7f31a8
--- /dev/null
+++ b/src/s0015_3sum.cpp
@@ -0,0 +1,79 @@
+#include "s0015_3sum.hpp"
+
+// 思路一：
+// 遍历一次，创建一个哈希表
+// 双重遍历，找 - nums[i] - nums[j] 是否存在于哈希表中
+// 去重的方式：
+// 对于每个找到的序列，先排序，如果不存在则添加到结果中。
+
+vector<vector<int>> Solution1::threeSum(vector<int>& nums) {
+  unordered_map<int, int> map;
+  vector<vector<int>> result;
+  int len = nums.size();
+  if (len == 0) {
+    return result;
+  }
+  for (int i{0}; i < len; i++) {
+    map[nums.at(i)] = i;
+  }
+  for (int i{0}; i < len; i++) {
+    for (int j{0}; j < len; j++) {
+      if (i == j) {
+        continue;
+      } else {
+        if (map.count(0 - nums.at(i) - nums.at(j)) != 0) {
+          if (i == map[0 - nums.at(i) - nums.at(j)] ||
+              j == map[0 - nums.at(i) - nums.at(j)]) {
+            continue;
+          } else {
+            vector<int> v{nums.at(i), nums.at(j), 0 - nums.at(i) - nums.at(j)};
+            sort(v.begin(), v.end());
+            // push if not exist
+            if (std::find(result.begin(), result.end(), v) == result.end()) {
+              result.push_back(v);
+            }
+          }
+        }
+      }
+    }
+  }
+  return result;
+}
+
+// 思路二：双指针
+// 先对整个数组排序，然后固定一个数 a ，然后 b, c 从两边往中间靠。
+vector<vector<int>> Solution2::threeSum(vector<int>& nums) {
+  int n = nums.size();
+  sort(nums.begin(), nums.end());
+  vector<vector<int>> ans;
+  // 枚举 a
+  for (int first = 0; first < n; ++first) {
+    // 需要和上一次枚举的数不相同
+    if (first > 0 && nums[first] == nums[first - 1]) {
+      continue;
+    }
+    // c 对应的指针初始指向数组的最右端
+    int third = n - 1;
+    int target = -nums[first];
+    // 枚举 b
+    for (int second = first + 1; second < n; ++second) {
+      // 需要和上一次枚举的数不相同
+      if (second > first + 1 && nums[second] == nums[second - 1]) {
+        continue;
+      }
+      // 需要保证 b 的指针在 c 的指针的左侧
+      while (second < third && nums[second] + nums[third] > target) {
+        --third;
+      }
+      // 如果指针重合，随着 b 后续的增加
+      // 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
+      if (second == third) {
+        break;
+      }
+      if (nums[second] + nums[third] == target) {
+        ans.push_back({nums[first], nums[second], nums[third]});
+      }
+    }
+  }
+  return ans;
+}
diff --git a/tests/s0015_3sum.cpp b/tests/s0015_3sum.cpp
new file mode 100644
index 0000000..34b4b76
--- /dev/null
+++ b/tests/s0015_3sum.cpp
@@ -0,0 +1,49 @@
+#include "s0015_3sum.hpp"
+
+#include <gtest/gtest.h>
+
+TEST(Problem15, Case1) {
+  vector<int> i{-1, 0, 1, 2, -1, -4};
+  vector<vector<int>> o1{{-1, -1, 2}, {-1, 0, 1}};
+  vector<vector<int>> o2{{-1, 0, 1}, {-1, -1, 2}};
+  Solution1 solution1;
+  Solution2 solution2;
+  EXPECT_TRUE(solution1.threeSum(i) == o1 || solution1.threeSum(i) == o2);
+  EXPECT_TRUE(solution2.threeSum(i) == o1 || solution2.threeSum(i) == o2);
+}
+
+TEST(Problem15, Case2) {
+  vector<int> i{0, 1, 1};
+  vector<vector<int>> o{};
+  Solution1 solution1;
+  Solution2 solution2;
+  EXPECT_EQ(solution1.threeSum(i), o);
+  EXPECT_EQ(solution2.threeSum(i), o);
+}
+
+TEST(Problem15, Case3) {
+  vector<int> i{0, 0, 0};
+  vector<vector<int>> o{{0, 0, 0}};
+  Solution1 solution1;
+  Solution2 solution2;
+  EXPECT_EQ(solution1.threeSum(i), o);
+  EXPECT_EQ(solution2.threeSum(i), o);
+}
+
+TEST(Problem15, Case4) {
+  vector<int> i{-1, 0, 1};
+  vector<vector<int>> o{{-1, 0, 1}};
+  Solution1 solution1;
+  Solution2 solution2;
+  EXPECT_EQ(solution1.threeSum(i), o);
+  EXPECT_EQ(solution2.threeSum(i), o);
+}
+
+TEST(Problem15, Case5) {
+  vector<int> i{1, 2, -2, -1};
+  vector<vector<int>> o{};
+  Solution1 solution1;
+  Solution2 solution2;
+  EXPECT_EQ(solution1.threeSum(i), o);
+  EXPECT_EQ(solution2.threeSum(i), o);
+}