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+# [151. Reverse Words in a String](https://leetcode.com/problems/reverse-words-in-a-string/)
+# 思路
+翻转字符串中的单词，题目中给了我们写特别说明，如果单词之间遇到多个空格，只能返回一个，而且首尾不能有单词.
+
+## 思路一
+我们新建一个初始化为空的结果字符串res, 从后往前遍历字符串s将每个单词不断拼接到res的尾部就可以了.
+
+## 思路二
+题目还要求对使用C的同学满足O(1)的复杂度, 那我们就只能在s上进行修改了.
+
+我们先整个字符串整体翻转一次，然后再分别翻转每一个单词，此时就能得到我们需要的结果了。
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    string reverseWords(string s) {
+        string res = "";
+        int low, high = s.size() - 1;
+        while(high >= 0){
+            while(high >= 0 && s[high] == ' ') high--;
+            if(high < 0) return res;
+            low = high - 1;
+            while(low >= 0 && s[low] != ' ') low--;
+            
+            if(res != "") res += " ";
+            res += s.substr(low+1, high - low);
+            high = low - 1;
+        }
+        return res;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+public:
+    string reverseWords(string s) {
+        reverse(s.begin(), s.end());
+        int len = 0; // 结果字符串的长度
+        for(int i = 0; i < s.size(); i++){
+            if(s[i] != ' '){
+                if(len != 0) s[len++] = ' ';
+                int j = i;
+                while(j < s.size() && s[j] != ' ') s[len++] = s[j++];
+                reverse(s.begin() + len - (j - i), s.begin() + len);
+                i = j;
+            }
+        }
+        s.resize(len); // 修改字符串长度
+        return s;
+    }
+};
+```