diff --git a/code/lc102.java b/code/lc102.java
index 8d63c77f3c3a9775b2104b5485857f727b078188..dd6d9c03f26604aa3874112f8ba30a06349ca3e3 100644
--- a/code/lc102.java
+++ b/code/lc102.java
@@ -23,7 +23,7 @@ public class lc102 {
         if(root==null)
             return res;
         qu.add(root);
-        while(!qu.isEmpty()){
+        while(!qu.isEmpty()){   //两个while
             int size = qu.size();
             List<Integer> temp = new ArrayList<>();
             while(size>0){
diff --git a/code/lc113.java b/code/lc113.java
index b8515e07238ee743c3f4bceef9301533d3435907..230aa48cfa4ea578c783a21f949f190f6072bab5 100644
--- a/code/lc113.java
+++ b/code/lc113.java
@@ -20,20 +20,20 @@ public class lc113 {
         }
     }
     public List<List<Integer>> pathSum(TreeNode root, int sum) {
-        List<List<Integer>> res = new ArrayList<>();
-        helper(res, root, sum, 0, new ArrayList<>());
+        List<List<Integer>> res = new ArrayList();
+        helper(res, new ArrayList(), root, sum);
         return res;
     }
-    public void helper(List<List<Integer>> res, TreeNode root, int sum, int curr, List<Integer> curr_ls) {
+    public void helper(List<List<Integer>> res, List<Integer> cur, TreeNode root, int sum){
         if(root==null) return;
-        curr_ls.add(root.val);
-        if(curr+root.val==sum && root.left==null && root.right==null) { //到叶子节点
-            res.add(new ArrayList<>(curr_ls));
-            curr_ls.remove(curr_ls.size()-1);
-            return;
+        cur.add(root.val);
+        if(root.left==null&&root.right==null&&root.val==sum){   //到叶子节点
+            res.add(new ArrayList(cur));
+        }else{
+            helper(res, cur, root.left, sum-root.val);
+            helper(res, cur, root.right, sum-root.val);
         }
-        helper(res, root.left, sum, curr+root.val, curr_ls);
-        helper(res, root.right, sum, curr+root.val, curr_ls);
-        curr_ls.remove(curr_ls.size()-1);
+        cur.remove(cur.size()-1);
+        return;
     }
 }
diff --git a/code/lc123.java b/code/lc123.java
index 7ff46543c34ba8a0c44c27a93ea9bf755c21ed6a..8fc5c2f7eb538660121eea5b145b852643327769 100644
--- a/code/lc123.java
+++ b/code/lc123.java
@@ -1,6 +1,6 @@
 package code;
 /*
- * 122. Best Time to Buy and Sell Stock III
+ * 123. Best Time to Buy and Sell Stock III
  * 题意：买卖股票最大利润，只能买卖2次
  * 难度：Hard
  * 分类：Array, Dynamic Programming
diff --git a/code/lc131.java b/code/lc131.java
index 8d048e169f0f239a3a84f8adc8e9309f6f70a9e1..165b6e1c5b804e779c53660ec3a3bcfc459036d3 100644
--- a/code/lc131.java
+++ b/code/lc131.java
@@ -10,6 +10,7 @@ import java.util.List;
  * 思路：典型回溯法，注意向res添加内容时要重新new一下
  * Tips： lc5, lc9, lc125, lc131, lc234, lc647
  *       lc39
+ *       lc132
  *       判断是否为回文的方法：
  *       1. 从中心往两边扩充，中心可能是一个字符，也可能是两个字符
  *       2. dp，利用之前计算的结果，只判断边缘两个字符是否相等，从后往前dp
diff --git a/code/lc132.java b/code/lc132.java
new file mode 100644
index 0000000000000000000000000000000000000000..c18d13065aab13f7ebe3f51181bc18ea5555e608
--- /dev/null
+++ b/code/lc132.java
@@ -0,0 +1,52 @@
+package code;
+import java.util.Arrays;
+
+/*
+ * 132. Palindrome Partitioning II
+ * 题意：最少切几刀，回文串
+ * 难度：Hard
+ * 分类：Dynamic Programming
+ * 思路：和lc300思路很像，只是多了判断。dp[i]表示从0到i这一段切法的最小值，每次遍历前边的结果，看能否接上。
+ *      可以把判断回文和dp两个合并，因为用dp的形式判断回文
+ * Tips：lc300
+ *      bingo!
+ */
+
+public class lc132 {
+    public int minCut(String s) {
+        int[] dp = new int[s.length()+1];
+        Arrays.fill(dp, s.length());    //fill最大值
+        dp[0] = 0;
+        for(int i=1; i<dp.length; i++){
+            for(int j=0; j<=i; j++){    //注意<=
+                if(dp[j]!=s.length() && isPalindrome(s.substring(j,i)))
+                    dp[i] = Math.min(dp[i], dp[j]+1);
+            }
+        }
+        return dp[s.length()]-1; //要-1 最后一刀是在末尾
+    }
+
+    public Boolean isPalindrome(String str){
+        if(new StringBuilder(str).reverse().toString().equals(str)) return true;   //reverse一下
+        else return false;
+    }
+
+    public int minCut2(String s) {
+        char[] c = s.toCharArray();
+        int n = c.length;
+        int[] cut = new int[n];
+        boolean[][] pal = new boolean[n][n];
+
+        for(int i = 0; i < n; i++) {
+            int min = i;
+            for(int j = 0; j <= i; j++) {
+                if(c[j] == c[i] && (j + 1 > i - 1 || pal[j + 1][i - 1])) {
+                    pal[j][i] = true;
+                    min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
+                }
+            }
+            cut[i] = min;
+        }
+        return cut[n - 1];
+    }
+}
diff --git a/code/lc139.java b/code/lc139.java
index 67ac6a61c0adb0d377bddb46367e11561a7d6218..1d767356017c46f1b10e9f7eb8bd9af88984fe87 100644
--- a/code/lc139.java
+++ b/code/lc139.java
@@ -6,8 +6,10 @@ package code;
  * 分类：Dynamic Programming
  * 思路：动态规划
  * Tips：巧妙的方法，防止了复杂的操作，通过遍历之前计算出来的结果
+ *       递归的方法本质和dp是一样的，记住用备忘录算法，把之前的结果记下来
  *       lc140
  */
+import java.util.HashMap;
 import java.util.List;
 
 public class lc139 {
@@ -22,4 +24,16 @@ public class lc139 {
         }
         return dp[s.length()];
     }
+
+    HashMap<String, Boolean> hm = new HashMap();
+    public boolean wordBreak2(String s, List<String> wordDict) {
+        if(hm.containsKey(s)) return hm.get(s);
+        if(s.length() == 0) return true;
+        Boolean flag = false;
+        for(String word:wordDict){
+            if(s.startsWith(word)) flag = flag||wordBreak(s.substring(word.length()), wordDict);    //注意函数 startsWith
+        }
+        hm.put(s, flag);
+        return flag;
+    }
 }
diff --git a/code/lc141.java b/code/lc141.java
index 1f8bf71edf569dace8539060faa1ee5d4cc03b67..971a82c3ca838c86d09052c6ee2483db79e48eea 100644
--- a/code/lc141.java
+++ b/code/lc141.java
@@ -14,15 +14,13 @@ public class lc141 {
         ListNode(int x) { val = x; }
     }
     public boolean hasCycle(ListNode head) {
-        if(head==null)
-            return false;
-        ListNode faster = head;
+        if(head==null||head.next==null) return false;
         ListNode slow = head;
-        while( faster.next!=null && faster.next.next!=null){    //注意判断条件，slow一定不等于null，不用判断了
+        ListNode fast = head.next;
+        while(fast!=null&&fast.next!=null){  //注意判断条件，slow一定不等于null，不用判断了
             slow = slow.next;
-            faster = faster.next.next;
-            if(slow==faster)
-                return true;
+            fast = fast.next.next;
+            if(fast==slow) return true;
         }
         return false;
     }
diff --git a/code/lc142.java b/code/lc142.java
index 389e108065181d6b2adabb95bab41a9ec59243c2..b7b8adfc7811b17d67be27c856461df0a53af000 100644
--- a/code/lc142.java
+++ b/code/lc142.java
@@ -32,4 +32,23 @@ public class lc142 {
         }
         return null;
     }
+
+    public ListNode detectCycle2(ListNode head) {
+        if(head==null||head.next==null) return null;
+        ListNode slow = head;
+        ListNode fast = head.next;
+        while(fast!=null&&fast.next!=null){
+            slow = slow.next;
+            fast = fast.next.next;
+            if(slow==fast){
+                slow = slow.next;
+                while(head!=slow){
+                    head = head.next;
+                    slow = slow.next;
+                }
+                return slow;
+            }
+        }
+        return null;
+    }
 }
diff --git a/code/lc148.java b/code/lc148.java
index 290e85ec78c8256a9e0191a6902cdd88b3e69a98..61b55e76c397356809c792f2543d305dc7ca8f7a 100644
--- a/code/lc148.java
+++ b/code/lc148.java
@@ -25,7 +25,7 @@ public class lc148 {
         if( head==null || head.next == null ){
             return head;
         }
-        ListNode slow = head;
+        ListNode slow = head;   //记一下
         ListNode fast = head.next;
         while( fast!=null && fast.next!=null ){   //把链表分成两半
             slow = slow.next;
diff --git a/code/lc207.java b/code/lc207.java
index 8e7f0757a65350fac90c3da47981871499c0e3f3..7a387b072f7dd8e073cafaf200e7744314512e5a 100644
--- a/code/lc207.java
+++ b/code/lc207.java
@@ -23,7 +23,7 @@ public class lc207 {
         for (int i = 0; i < prerequisites.length ; i++) {
             int node1 = prerequisites[i][0];
             int node2 = prerequisites[i][1];
-            graph[node2][node1] = 1;
+            graph[node2][node1] = 1;    //存下邻接矩阵，方便后续查找是否有边
             indegree[node1]++;  //存下入度，入度为0时，表示该课程可以修
         }
         Stack<Integer> st = new Stack();
diff --git a/code/lc226.java b/code/lc226.java
index 30443bb56f88da6462866b4a9bd956f81b8e4be8..9524a5b1c20b8019cc06f629ea6ed56798c33604 100644
--- a/code/lc226.java
+++ b/code/lc226.java
@@ -19,8 +19,7 @@ public class lc226 {
     }
     public TreeNode invertTree(TreeNode root) {
         //递归
-        if(root==null)
-            return null;
+        if(root==null) return null;
         TreeNode temp = root.left;
         root.left = root.right;
         root.right = temp;
diff --git a/code/lc300.java b/code/lc300.java
index 1003cf49c94e97eba2af2aee4e36946cd87173e2..daefa73054ee8c13f912bb2619137a184550bfbc 100644
--- a/code/lc300.java
+++ b/code/lc300.java
@@ -6,6 +6,7 @@ package code;
  * 分类：Binary Search, Dynamic Programming
  * 思路：基本的思路是dp[i]记录以nums[i]结尾的最长长度，每次遍历 dp[i] 得到dp[i+1]，复杂度为O(n^2)。最优的解法是O(nlgn)，dp[i]是递增的数组，每次插入时二分查找是lgn。
  * Tips：经典题目，记一下
+ *      lc132
  */
 import java.util.Arrays;
 
diff --git a/code/lc309.java b/code/lc309.java
index a7d9b08cbe49eba535caaad5e206b377e0b65406..3a0ce768a934034b5b8aadce7d2e3edc283a9f0d 100644
--- a/code/lc309.java
+++ b/code/lc309.java
@@ -27,4 +27,18 @@ public class lc309 {
         }
         return Math.max(s,b);
     }
+
+    public int maxProfit2(int[] prices) {
+        if(prices.length==0) return 0;
+        int[] sell = new int[prices.length];
+        int[] buy = new int[prices.length];
+        sell[0] = 0;
+        buy[0] = -prices[0];
+        for(int i=1; i<prices.length; i++){
+            sell[i] = Math.max(sell[i-1], buy[i-1]+prices[i]);
+            if(i==1) buy[i] = Math.max(buy[i-1], 0-prices[i]);
+            else buy[i] = Math.max(buy[i-1], sell[i-2]-prices[i]);
+        }
+        return Math.max(sell[prices.length-1],buy[prices.length-1]);
+    }
 }
diff --git a/code/lc31.java b/code/lc31.java
index 656bfc49a8b48be2a05316b6e59d3ded47a6de44..9444e5102f9d5840de36adc4581d3075a498bcf2 100644
--- a/code/lc31.java
+++ b/code/lc31.java
@@ -4,12 +4,12 @@ package code;
  * 题意：找出排列组合的下一个排列
  * 难度：Medium
  * 分类：Array
- * 思路：从后往前找第一个变小的数x，从这个数之后的数中找出第一个比x大的数，交换，再把之后的数逆序即可
+ * 思路：从后往前找第一个变小的数x，从后往前找出比第一个x大的数，交换，再把之后的数逆序即可
  * Tips：很典型的排列组合题，思路方法记忆一下。注意比较时是否有=。
  */
 public class lc31 {
     public static void main(String[] args) {
-        int[] nums = {2,3,1,3,3};
+        int[] nums = {1,2,3};
         nextPermutation(nums);
         for (int i:nums){
             System.out.println(i);
@@ -19,27 +19,25 @@ public class lc31 {
     public static void nextPermutation(int[] nums) {
         int ptr = nums.length-1;
 
-        //从后往前找第一个变小的数x
+        //从后往前找第一个变小的数x 从后往前找出比第一个x大的数
         while(ptr>0&&nums[ptr-1]>=nums[ptr]){// 注意是 >= {5,1,1} , 等于--
             ptr--;
         }
-
-        if(ptr!=0){
-            //从这个数之后的数中找出第一个比x大的数
-            int n = nums[ptr];
-            int ptr2 = ptr;
-            for(int i=ptr+1; i<nums.length; i++){   //这不用这么麻烦，后边的数有序的，这可以简化
-                if( nums[i]>nums[ptr-1] && nums[i]<=n ) {//注意 <= {2,3,1,3,3}
-                    n = nums[i];
-                    ptr2 = i;
-                }
+        ptr--;
+        if(ptr!=-1){
+            //从后往前，找比
+            int val = nums[ptr];
+            int ptr2 = nums.length-1;
+            while(ptr2>ptr){
+                if(nums[ptr2]>nums[ptr]) break;
+                ptr2--;
             }
-            nums[ptr2] = nums[ptr-1];
-            nums[ptr-1] = n;
+            nums[ptr] = nums[ptr2];
+            nums[ptr2] = val;
         }
 
         //把之后的数逆序
-        ReverseNums(nums,ptr,nums.length-1);
+        ReverseNums(nums,ptr+1,nums.length-1);  //+1，不包含ptr那个位置
     }
     public static void ReverseNums(int[] nums, int start, int end){
         int l = end+start;
diff --git a/code/lc32.java b/code/lc32.java
index 041752ea35056bb5f64970341b96a2a9a7af0eb0..d61ed8bf3fabb43902f8f1580caf5ca421016057 100644
--- a/code/lc32.java
+++ b/code/lc32.java
@@ -28,7 +28,7 @@ public class lc32 {
             else{
                 if(s.charAt(i-2)=='('){ // 这种情况：(())()
                     dp[i] = dp[i-2]+2;
-                }else if(i-2-dp[i-1]>=0 && s.charAt(i-2-dp[i-1])=='('){ // 这种情况：()(())
+                }else if(i-2-dp[i-1]>=0 && s.charAt(i-2-dp[i-1])=='('){ // 这种情况：()(()) , 第一个判断是判断索引是否合法
                     dp[i] = dp[i-1] + dp[i-2-dp[i-1]]+2;
                 }
             }
@@ -49,7 +49,7 @@ public class lc32 {
                 st.push(i);
             else if(ch==')'){
                 st.pop();
-                if(st.isEmpty())
+                if(st.isEmpty())    //截断一下
                     st.push(i);
                 res = Math.max(res,i-st.peek());
             }
diff --git a/code/lc416.java b/code/lc416.java
index a618c4020b2131e13cd7818e2c3c2bd104fd6003..a595529ed199273dfc09b33ea175049b7ab8a6cf 100644
--- a/code/lc416.java
+++ b/code/lc416.java
@@ -69,5 +69,21 @@ public class lc416 {
         return dp[volumn];
     }
 
+    public boolean canPartition3(int[] nums) {
+        int sum = 0;
+        for(int i: nums) sum+=i;
+        if(sum%2==1) return false;
+        sum = sum/2;
+        boolean[][] dp = new boolean[nums.length+1][sum+1];
+        for(int i =0; i<=nums.length; i++) dp[i][0]=true;   //注意赋值0
+        for(int i=1; i<=nums.length; i++){
+            for(int j=1; j<=sum; j++){
+                dp[i][j] = dp[i-1][j];
+                if(j>=nums[i-1]) dp[i][j]= dp[i][j] || dp[i-1][j-nums[i-1]];   //注意判断index是否合法
+            }
+        }
+        return dp[nums.length][sum];
+    }
+
 
 }
diff --git a/code/lc76.java b/code/lc76.java
index 6d075a0cab10818285e9fdc1ade182d03b95d03f..6b8c407f4d398566e91ce4464e02990762e04ccb 100644
--- a/code/lc76.java
+++ b/code/lc76.java
@@ -17,7 +17,7 @@ public class lc76 {
     }
     public static String minWindow(String s, String t) {
         HashMap<Character,Integer> mp = new HashMap();
-        for (int i = 0; i < t.length() ; i++) { //统计每个字符出现的个数
+        for (int i = 0; i < t.length() ; i++) { // 统计每个字符出现的个数
             char ch = t.charAt(i);
             if(mp.containsKey(ch))
                 mp.put(ch,mp.get(ch)+1);
@@ -37,7 +37,7 @@ public class lc76 {
                 if(mp.get(ch_r)>=0) // <0说明重复了
                     count++;
             }
-            while(count==t.length()){//右移左指针
+            while(count==t.length()){//右移左指针，注意这个判定条件
                 if(right-left+1<res_len){ //更新结果
                     res_left = left;
                     res_len = right-left+1;
@@ -56,4 +56,39 @@ public class lc76 {
             return "";
         return s.substring(res_left,res_left+res_len);
     }
+
+    public String minWindow2(String s, String t) {
+        HashMap<Character, Integer> hm = new HashMap();
+        char[] t_arr = t.toCharArray();
+        for(int i=0; i<t_arr.length; i++){
+            hm.put(t_arr[i], hm.getOrDefault(t_arr[i], 0)+1);
+        }
+        int left = 0;
+        int right = 0;
+        int count = t.length();
+        char[] s_arr = s.toCharArray();
+        int res = Integer.MAX_VALUE;
+        int res_left = 0;
+        int res_right = 0;
+        while(right<s.length()){    //一共两个while
+            if(hm.containsKey(s_arr[right])){
+                hm.put(s_arr[right], hm.get(s_arr[right])-1);
+                if(hm.get(s_arr[right])>=0) count--;    //别忘了这一层的判断
+            }
+            right++;    //在if外边
+            while(count==0){    //注意判断条件
+                if(right-left+1<res){   //记录结果
+                    res = right-left+1;
+                    res_left = left;
+                    res_right = right;
+                }
+                if(hm.containsKey(s_arr[left])){
+                    hm.put(s_arr[left], hm.get(s_arr[left])+1);
+                    if(hm.get(s_arr[left])>0) count++;
+                }
+                left++;
+            }
+        }
+        return s.substring(res_left, res_right);
+    }
 }
diff --git a/code/lc94.java b/code/lc94.java
index 1f93c8191c4a68c7f9a53656708aad0a34f75f63..1bd956e42ce997cd7bfb89fd63921564e0b6071e 100644
--- a/code/lc94.java
+++ b/code/lc94.java
@@ -5,7 +5,7 @@ package code;
  * 难度：Medium
  * 分类：HashTable, Stack, Tree
  * 思路：左节点依次入栈二叉树中序遍历
- * Tips：和lc144前序,lc145后序一起看
+ * Tips：和lc144前序,lc145后序一起看, lc102层次遍历
  */
 import java.util.ArrayList;
 import java.util.List;
diff --git a/code/lc978.java b/code/lc978.java
index 1cdb748cf6a4d0dcc0ef65750c4067c00eef6745..35653a1b0968173a8a0c2b42a034bedba482d93f 100644
--- a/code/lc978.java
+++ b/code/lc978.java
@@ -28,7 +28,7 @@ public class lc978 {
 
     public int maxTurbulenceSize2(int[] A) {
         int[] arr = new int[A.length-1];
-        for(int i=1; i<A.length; i++){
+        for(int i=1; i<A.length; i++){  //转换成0,1数组
             if(A[i]==A[i-1]) arr[i-1] = 0;
             else if(A[i]>A[i-1]) arr[i-1] = 1;
             else arr[i-1] = -1;