diff --git a/README.md b/README.md
index 0fc650a9..e5b5c53e 100644
--- a/README.md
+++ b/README.md
@@ -23,9 +23,11 @@ LeetCode 最强题解（持续更新中）:
 |[0219.存在重复元素II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0219.存在重复元素II.md) | 哈希表 |简单| **哈希** |
 |[0237.删除链表中的节点](https://github.com/youngyangyang04/leetcode/blob/master/problems/0237.删除链表中的节点.md) |链表 |简单| **原链表移除** **添加虚拟节点** 递归|
 |[0242.有效的字母异位词](https://github.com/youngyangyang04/leetcode/blob/master/problems/0242.有效的字母异位词.md) |哈希表 |简单| **哈希**|
+|[0344.反转字符串](https://github.com/youngyangyang04/leetcode/blob/master/problems/0344.反转字符串.md) |字符串 |简单| **双指针**|
 |[0349.两个数组的交集](https://github.com/youngyangyang04/leetcode/blob/master/problems/0349.两个数组的交集.md) |哈希表 |简单|**哈希**|
 |[0350.两个数组的交集II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0350.两个数组的交集II.md) |哈希表 |简单|**哈希**|
 |[0383.赎金信](https://github.com/youngyangyang04/leetcode/blob/master/problems/0383.赎金信.md) |数组 |简单|**暴力** **字典计数** **哈希**|
+|[0434.字符串中的单词数](https://github.com/youngyangyang04/leetcode/blob/master/problems/0434.字符串中的单词数.md) |字符串 |简单|**模拟**|
 |[0454.四数相加II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0454.四数相加II.md) |哈希表 |中等| **哈希**|
 |[0575.分糖果.md](https://github.com/youngyangyang04/leetcode/blob/master/problems/0575.分糖果.md) |哈希表 |简单|**哈希**|
 |[0705.设计哈希集合.md](https://github.com/youngyangyang04/leetcode/blob/master/problems/0705.设计哈希集合.md) |哈希表 |简单|**模拟**|
diff --git a/problems/0434.字符串中的单词数.md b/problems/0434.字符串中的单词数.md
new file mode 100644
index 00000000..bd4fd4f2
--- /dev/null
+++ b/problems/0434.字符串中的单词数.md
@@ -0,0 +1,28 @@
+## 题目地址 
+https://leetcode-cn.com/problems/number-of-segments-in-a-string/
+
+## 思路 
+
+可以把问题抽象为 遇到s[i]是空格，s[i+1]不是空格，就统计一个单词， 然后特别处理一下第一个字符不是空格的情况
+
+## C++代码
+
+```
+class Solution {
+public:
+    int countSegments(string s) {
+        int count = 0;
+        for (int i = 0; i < s.size(); i++) {
+            // 第一个字符不是空格的情况
+            if (i == 0 && s[i] != ' ') {
+                count++;
+            }
+            // 只要s[i]是空格，s[i+1]不是空格，count就加1
+            if (i + 1 < s.size() && s[i] == ' ' && s[i + 1] != ' ') {
+                count++;
+            }
+        }
+        return count;
+    }
+};
+```