diff --git a/README.md b/README.md
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+++ b/README.md
@@ -70,6 +70,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 69 |[Sqrt(x)](https://leetcode.com/problems/sqrtx)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/69.%20Sqrt(x).md)|Easy|  |
 | 70 |[Climbing Stairs](https://leetcode.com/problems/climbing-stairs)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/70.%20Climbing%20Stairs.md)|Easy|  |
 | 71 |[Simplify Path](https://leetcode.com/problems/simplify-path/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/71.%20Simplify%20Path.md)|Medium|  |
+| 72 |[Edit Distance](https://leetcode.com/problems/edit-distance/)|[C++](solutions/72.%20Edit%20Distance.md)|Hard|  |
 | 73 |[Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/73.%20Set%20Matrix%20Zeroes.md)|Medium|  |
 | 74 |[Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/74.%20Search%20a%202D%20Matrix.md)|Medium|  |
 | 75 |[Sort Colors](https://leetcode.com/problems/sort-colors/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/75.%20Sort%20Colors.md)|Medium|  |
diff --git a/solutions/72. Edit Distance.md b/solutions/72. Edit Distance.md
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+# [72. Edit Distance](https://leetcode.com/problems/edit-distance/)
+
+# 思路
+
+求编辑距离：给定两个字符串word1和word2，求word1经过几次操作能变成word2。
+
+根据经验，这种求最值的两个字符串类的问题基本都可以用动归（因为整个问题可以分解成多个子问题求解而且子问题之间有重复），这里也不例外。我们定义
+```
+dp[i][j] 表示字符串word1[0,1,...,i-1]和word1[0,1,...,j-1]的编辑距离
+```
+需要仔细考虑一下初始情况，即当两个串中有一个空串时编辑距离就为另一个串的长度，所以我们可以按照下面初始化：
+
+``` C++
+dp[0][0] = 0; // 两个空串
+for(int i = 1; i <= n1; i++) dp[i][0] = i; // word2是空串
+for(int i = 1; i <= n2; i++) dp[0][i] = i; // word1是空串
+```
+
+状态转移时有两种情况：
+1. `word1[i-1] == word2[j-1]`，那么此时很简单`dp[i][j] = dp[i-1][j-1], dp[i][j]`；
+2. 否则，我们需要分别对word1末尾进行插入、删除和替换，取三者最小结果即可，即`dp[i][j] = 1 + min(dp[i-1][j-1], min(dp[i][j-1], dp[i-1][j]))`。
+
+可见状态数组里的元素`dp[i][[j]`只与左上三个方向的值（`dp[i-1][j-1], dp[i][j-1], dp[i-1][j]`）有关，所以可用滚动数组优化空间至线性。
+
+时间复杂度O(mn)，空间复杂度O(mn)（可优化至O(n)）
+
+# C++
+``` C++
+class Solution {
+public:
+    int minDistance(string word1, string word2) {
+        int n1 = word1.size(), n2 = word2.size();
+        
+        vector<vector<int>>dp(n1 + 1, vector<int>(n2 + 1, 0));
+        for(int i = 1; i <= n1; i++) dp[i][0] = i;
+        for(int i = 1; i <= n2; i++) dp[0][i] = i;
+        
+        for(int i = 1; i <= n1; i++){
+            for(int j = 1; j <= n2; j++){
+                if(word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1], dp[i][j];
+                else dp[i][j] = 1 + min(dp[i-1][j-1], min(dp[i][j-1], dp[i-1][j]));
+            }
+        }
+        return dp[n1][n2];
+    }
+};
+```
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