diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index 6a5d33a9..48412ed4 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -61,6 +61,8 @@ public:
     vector<vector<int>> threeSum(vector<int>& nums) {
         vector<vector<int>> result;
         sort(nums.begin(), nums.end());
+        // 找出a + b + c = 0
+        // a = nums[i], b = nums[left], c = nums[right]
         for (int i = 0; i < nums.size(); i++) {
             // 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
             if (nums[i] > 0) {
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index 425cb805..d896abc1 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -1,4 +1,3 @@
-
 ## 题目地址 
 
 https://leetcode-cn.com/problems/valid-anagram/
diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md
new file mode 100644
index 00000000..fc0a779e
--- /dev/null
+++ b/problems/0344.反转字符串.md
@@ -0,0 +1,19 @@
+## 题目地址 
+https://leetcode-cn.com/problems/reverse-string/
+
+## 思路
+
+遍历数组的前一半，同时和后一半做交换就可以了
+
+## 代码
+
+```
+class Solution {
+public:
+    void reverseString(vector<char>& s) {
+        for (int i = 0, j = s.size() - 1; i < s.size()/2; i++, j--) {
+            swap(s[i],s[j]);
+        }
+    }
+};
+```