diff --git a/problems/0053.最大子序和（动态规划）.md b/problems/0053.最大子序和（动态规划）.md
index 37de9bbe..703e1dd6 100644
--- a/problems/0053.最大子序和（动态规划）.md
+++ b/problems/0053.最大子序和（动态规划）.md
@@ -174,6 +174,7 @@ const maxSubArray = nums => {
     // 数组长度，dp初始化
     const len = nums.length;
     let dp = new Array(len).fill(0);
+    dp[0] = nums[0];
     // 最大值初始化为dp[0]
     let max = dp[0];
     for (let i = 1; i < len; i++) {
diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md
index 1653a81a..e04cb173 100644
--- a/problems/0139.单词拆分.md
+++ b/problems/0139.单词拆分.md
@@ -89,27 +89,26 @@ class Solution {
 private:
     bool backtracking (const string& s,
             const unordered_set<string>& wordSet,
-            vector<int>& memory,
+            vector<bool>& memory,
             int startIndex) {
         if (startIndex >= s.size()) {
             return true;
         }
         // 如果memory[startIndex]不是初始值了，直接使用memory[startIndex]的结果
-        if (memory[startIndex] != -1) return memory[startIndex];
+        if (!memory[startIndex]) return memory[startIndex];
         for (int i = startIndex; i < s.size(); i++) {
             string word = s.substr(startIndex, i - startIndex + 1);
             if (wordSet.find(word) != wordSet.end() && backtracking(s, wordSet, memory, i + 1)) {
-                memory[startIndex] = 1; // 记录以startIndex开始的子串是可以被拆分的
                 return true;
             }
         }
-        memory[startIndex] = 0; // 记录以startIndex开始的子串是不可以被拆分的
+        memory[startIndex] = false; // 记录以startIndex开始的子串是不可以被拆分的
         return false;
     }
 public:
     bool wordBreak(string s, vector<string>& wordDict) {
         unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
-        vector<int> memory(s.size(), -1); // -1 表示初始化状态
+        vector<bool> memory(s.size(), 1); // -1 表示初始化状态
         return backtracking(s, wordSet, memory, 0);
     }
 };
diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md
index 6213aeaa..ca5fba77 100644
--- a/problems/0236.二叉树的最近公共祖先.md
+++ b/problems/0236.二叉树的最近公共祖先.md
@@ -325,6 +325,20 @@ var lowestCommonAncestor = function(root, p, q) {
 };
 ```
 
+## TypeScript
+
+```typescript
+function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: TreeNode | null): TreeNode | null {
+    if (root === null || root === p || root === q) return root;
+    const left = lowestCommonAncestor(root.left, p, q);
+    const right = lowestCommonAncestor(root.right, p, q);
+    if (left !== null && right !== null) return root;
+    if (left !== null) return left;
+    if (right !== null) return right;
+    return null;
+};
+```
+
 
 
 -----------------------
diff --git a/problems/0300.最长上升子序列.md b/problems/0300.最长上升子序列.md
index ed61a30e..dfdd5125 100644
--- a/problems/0300.最长上升子序列.md
+++ b/problems/0300.最长上升子序列.md
@@ -37,7 +37,7 @@
 
 1. dp[i]的定义
 
-**dp[i]表示i之前包括i的最长上升子序列的长度**。
+**dp[i]表示i之前包括i的以nums[i]结尾最长上升子序列的长度**
 
 2. 状态转移方程
 
diff --git a/problems/0738.单调递增的数字.md b/problems/0738.单调递增的数字.md
index e670bb31..c8ce8a2b 100644
--- a/problems/0738.单调递增的数字.md
+++ b/problems/0738.单调递增的数字.md
@@ -148,23 +148,19 @@ java版本1中创建了String数组，多次使用Integer.parseInt了方法，
 版本2
 class Solution {
     public int monotoneIncreasingDigits(int n) {
-        if (n==0)return 0;
-        char[] chars= Integer.toString(n).toCharArray();
-        int start=Integer.MAX_VALUE;//start初始值设为最大值，这是为了防止当数字本身是单调递增时，没有一位数字需要改成9的情况
-        for (int i=chars.length-1;i>0;i--){
-            if (chars[i]<chars[i-1]){
-                chars[i-1]--;
-                start=i;
+        String s = String.valueOf(n);
+        char[] chars = s.toCharArray();
+        int start = s.length();
+        for (int i = s.length() - 2; i >= 0; i--) {
+            if (chars[i] > chars[i + 1]) {
+                chars[i]--;
+                start = i+1;
             }
         }
-        StringBuilder res=new StringBuilder();
-        for (int i=0;i<chars.length;i++){
-            if (chars[i]=='0'&&i==0)continue;//防止出现09这样的情况
-            if (i>=start){
-                res.append('9');
-            }else res.append(chars[i]);
+        for (int i = start; i < s.length(); i++) {
+            chars[i] = '9';
         }
-        return Integer.parseInt(res.toString());
+        return Integer.parseInt(String.valueOf(chars));
     }
 }
 ```