diff --git a/code/lc15.java b/code/lc15.java
index 29a9fc25353837d3ed8302f4493530eb3fd5d402..781fc428f8342932e634c1b34a65d5eba5d11705 100644
--- a/code/lc15.java
+++ b/code/lc15.java
@@ -5,6 +5,7 @@ package code;
  * 难度：Medium
  * 分类：Array, Two Pointers
  * 注意：如何避免 List 重复元素
+ * Tips：lc15, lc16, lc923
  */
 import java.util.*;
 
diff --git a/code/lc16.java b/code/lc16.java
index b18f4130094fc5eb5ff6997156a755a4d771063a..7747291deb8dcc311d61df1a37a939fffa1b12ea 100644
--- a/code/lc16.java
+++ b/code/lc16.java
@@ -7,7 +7,7 @@ import java.util.Arrays;
  * 难度：Medium
  * 分类：Array, Two Pointers
  * 思路：3sum的思路，每次记下最接近的res即可
- * Tips：
+ * Tips：lc15, lc16, lc923
  */
 public class lc16 {
     public int threeSumClosest(int[] nums, int target) {
diff --git a/code/lc53.java b/code/lc53.java
index e134b12b7718d9c4ee68deccd21f647330ee0dcd..b73be9908a69528428240c6fc51245ebb1b85446 100644
--- a/code/lc53.java
+++ b/code/lc53.java
@@ -5,6 +5,7 @@ package code;
  * 难度：Easy
  * 分类：Array, Divide and Conquer, Dynamic Programming
  * 注意：分治方法如何进行merge，merge时，必须包含mid元素，因为是连续子序列
+ *      lc978
  */
 public class lc53 {
     public static void main(String[] args) {
diff --git a/code/lc560.java b/code/lc560.java
index f5eb8449c5670373f5cb22d54faf8f0c9268dd53..e7f6d158f0fca63a6e43bef1658a9059804f14ec 100644
--- a/code/lc560.java
+++ b/code/lc560.java
@@ -4,7 +4,7 @@ import java.util.HashMap;
 
 /*
  * 560. Subarray Sum Equals K
- * 题意：子数组的和等于k的个数
+ * 题意：连续子数组的和等于k的个数
  * 难度：Medium
  * 分类：Array, Hash Table
  * 思路：求出累加和存在hashmap中，如果当前hashmap中存在sum-k，那么就是一个解
diff --git a/code/lc921.java b/code/lc921.java
new file mode 100644
index 0000000000000000000000000000000000000000..88d463a89db7dd3cf6ab497c6816822f0b635f29
--- /dev/null
+++ b/code/lc921.java
@@ -0,0 +1,25 @@
+package code;
+
+import java.util.Stack;
+/*
+ * 921. Minimum Add to Make Parentheses Valid
+ * 题意：最少添加符号个数，使得表达式有效
+ * 难度：Medium
+ * 分类：Stack, Greedy
+ * 思路：匹配的都出栈，最后剩下的栈中没匹配的个数，就是需要添加的个数
+ * Tips：
+ */
+public class lc921 {
+    public int minAddToMakeValid(String S) {
+        char[] ch_arr = S.toCharArray();
+        Stack<Character> st = new Stack<>();
+        for (int i = 0; i < ch_arr.length ; i++) {
+            if(ch_arr[i]==')' && !st.isEmpty() && st.peek()=='('){
+                st.pop();
+            }else{
+                st.push(ch_arr[i]);
+            }
+        }
+        return st.size();
+    }
+}
diff --git a/code/lc922.java b/code/lc922.java
new file mode 100644
index 0000000000000000000000000000000000000000..b3022ebfd09009368b9de069071466e09be8ad8e
--- /dev/null
+++ b/code/lc922.java
@@ -0,0 +1,38 @@
+package code;
+/*
+ * 922. Sort Array By Parity II
+ * 题意：奇数位置上都是奇数，偶数位置上都是偶数
+ * 难度：Easy
+ * 分类：Array, Sort
+ * 思路：题看似简单，但最优的方法并不好写
+ * Tips：
+ */
+public class lc922 {
+    public int[] sortArrayByParityII(int[] A) { //O(N^2)
+        for (int i = 0; i < A.length ; i++) {
+            if( A[i]%2==i%2 ) continue;
+            for (int j = i+1; j < A.length ; j++) {
+                if(A[i]%2!=A[j]%2) {
+                    int temp = A[i];
+                    A[i] = A[j];
+                    A[j] = temp;
+                    break;
+                }
+            }
+        }
+        return A;
+    }
+
+    public int[] sortArrayByParityII2(int[] A) { //很巧妙的方法  O(N)时间， O(1)空间
+        int cur = 1;    //奇数位置的位置
+        for (int i = 0; i < A.length ; i+=2) { //这里+2 只判断偶数位置上是否符合
+            if(A[i]%2==1){
+                while(A[cur]%2==1) cur+=2;
+                int temp = A[cur];
+                A[cur] = A[i];
+                A[i] = temp;
+            }
+        }
+        return A;
+    }
+}
diff --git a/code/lc923.java b/code/lc923.java
new file mode 100644
index 0000000000000000000000000000000000000000..44c0a32b097d1e1886091c155c33b9b8d581daca
--- /dev/null
+++ b/code/lc923.java
@@ -0,0 +1,73 @@
+package code;
+import java.util.Arrays;
+import java.util.HashMap;
+
+/*
+ * 923. 3Sum With Multiplicity
+ * 题意：3Sum有几种？数组中有重复数字
+ * 难度：Medium
+ * 分类：Two Pointers
+ * 思路：由于本题只要求给出多少种Int值，所以不一定非要用3Sum的思路，有很多更简答的方法
+ *      3种思路
+ * Tips：
+ */
+public class lc923 {
+    public int threeSumMulti(int[] A, int target) {
+        int res = 0;
+        HashMap<Integer, Integer> hm = new HashMap();
+        for (int i = 0; i < A.length ; i++) {
+            res = (res+hm.getOrDefault(target-A[i],0))%1000000007;  // i之前的数字，两个组合起来，是否 == target-A[i]
+            for (int j = 0; j < i ; j++) {
+                hm.put(A[i]+A[j], hm.getOrDefault(A[i]+A[j], 0)+1);
+            }
+        }
+        return res;
+    }
+    public int threeSumMulti2(int[] A, int target) {    //3Sum的思路
+        int res = 0;
+        Arrays.sort(A);
+        for (int i = 0; i < A.length-2 ; i++) {
+            int left = i+1, right = A.length-1;
+            while(left<right){
+                if(A[i]+A[left]+A[right]<target) left++;
+                else if(A[i]+A[left]+A[right]>target) right--;
+                else if(A[left]==A[right]) {    //如果相等，则直接 C N 取 2，计算出来，然后break
+                    res = (res + (right-left)*(right-left+1)/2)%1000000007;
+                    break;  //不用继续移动指针了
+                } else {
+                    int leftcount = 1, rightcount = 1;
+                    while(A[left]==A[left+1]) {
+                        left++;
+                        leftcount++;
+                    }
+                    while(A[right]==A[right-1]) {
+                        right--;
+                        rightcount++;
+                    }
+                    res = (res + leftcount*rightcount)%1000000007;
+                    left++;     //别忘了，最后还要操作一下
+                    right--;
+                }
+            }
+        }
+        return res;
+    }
+
+    public int threeSumMulti3(int[] A, int target) {    //直接数学计算
+        long[] c = new long[101];
+        for (int a : A) c[a]++;
+        long res = 0;
+        for (int i = 0; i <= 100; i++)  // 题目给了值不超过100
+            for (int j = i; j <= 100; j++) {
+                int k = target - i - j;
+                if (k > 100 || k < 0) continue;
+                if (i == j && j == k)
+                    res += c[i] * (c[i] - 1) * (c[i] - 2) / 6;
+                else if (i == j && j != k)
+                    res += c[i] * (c[i] - 1) / 2 * c[k];
+                else if (j < k)
+                    res += c[i] * c[j] * c[k];
+            }
+        return (int)(res % (1e9 + 7));
+    }
+}
diff --git a/code/lc978.java b/code/lc978.java
new file mode 100644
index 0000000000000000000000000000000000000000..b51ba2fe9b55b74ced3569dd036884d8af2d50af
--- /dev/null
+++ b/code/lc978.java
@@ -0,0 +1,28 @@
+package code;
+/*
+ * 978. Longest Turbulent Subarray
+ * 题意：小，大，小，大 这种最大长度是多少
+ * 难度：Medium
+ * 分类：Array, Dynamic Porgramming, Sliding Windows
+ * 思路：O(n) 遍历一遍就能解决，不用dp
+ * Tips：lc53
+ */
+public class lc978 {
+    public int maxTurbulenceSize(int[] A) {
+        int inc = 1, dec = 1, result = 1;
+        for (int i = 1; i < A.length; i++) {
+            if (A[i] < A[i - 1]) {  // +1 并且重置另一个统计量
+                dec = inc + 1;
+                inc = 1;
+            } else if (A[i] > A[i - 1]) {
+                inc = dec + 1;
+                dec = 1;
+            } else {
+                inc = 1;
+                dec = 1;
+            }
+            result = Math.max(result, Math.max(dec, inc));
+        }
+        return result;
+    }
+}
diff --git a/readme.md b/readme.md
index 04d96b2f598e1ff9dbfb552c4c18644104761775..6de0c759be2d484ff1ce401ca1f543501fee7eaa 100644
--- a/readme.md
+++ b/readme.md
@@ -98,6 +98,7 @@ LeetCode 指南
 | 104 [Java](./code/lc104.java)
 | 105 [Java](./code/lc105.java)
 | 108 [Java](./code/lc108.java)
+| 112 [Java](./code/lc112.java)
 | 113 [Java](./code/lc113.java)
 | 114 [Java](./code/lc114.java)
 | 116 [Java](./code/lc116.java)
@@ -111,6 +112,7 @@ LeetCode 指南
 | 125 [Java](./code/lc125.java)
 | 127 [Java](./code/lc127.java)
 | 128 [Java](./code/lc128.java)
+| 129 [Java](./code/lc129.java)
 | 130 [Java](./code/lc130.java)
 | 131 [Java](./code/lc131.java)
 | 134 [Java](./code/lc134.java)
@@ -126,6 +128,7 @@ LeetCode 指南
 | 148 [Java](./code/lc148.java)
 | 149 [Java](./code/lc149.java)
 | 150 [Java](./code/lc150.java)
+| 151 [Java](./code/lc151.java)
 | 152 [Java](./code/lc152.java)
 | 155 [Java](./code/lc155.java)
 | 160 [Java](./code/lc160.java)