From f39d5a110c19963037374a3f1ea253b30f08becb Mon Sep 17 00:00:00 2001
From: erdengk <15596570256@163.com>
Date: Wed, 16 Feb 2022 16:23:39 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E9=83=A8=E5=88=86?=
 =?UTF-8?q?=E9=94=99=E5=AD=97=20&=20=E4=BC=98=E5=8C=96=E5=8F=98=E9=87=8F?=
 =?UTF-8?q?=E5=90=8D=E5=91=BD=E5=90=8D?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 ...序与后序遍历序列构造二叉树.md |  4 ++--
 problems/0860.柠檬水找零.md              | 20 +++++++++----------
 2 files changed, 12 insertions(+), 12 deletions(-)

diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index 30ec684d..cb1d75d9 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -89,9 +89,9 @@ TreeNode* traversal (vector<int>& inorder, vector<int>& postorder) {
 
 **难点大家应该发现了，就是如何切割，以及边界值找不好很容易乱套。**
 
-此时应该注意确定切割的标准，是左闭右开，还有左开又闭，还是左闭又闭，这个就是不变量，要在递归中保持这个不变量。
+此时应该注意确定切割的标准，是左闭右开，还有左开右闭，还是左闭右闭，这个就是不变量，要在递归中保持这个不变量。
 
-**在切割的过程中会产生四个区间，把握不好不变量的话，一会左闭右开，一会左闭又闭，必然乱套！**
+**在切割的过程中会产生四个区间，把握不好不变量的话，一会左闭右开，一会左闭右闭，必然乱套！**
 
 我在[数组：每次遇到二分法，都是一看就会，一写就废](https://programmercarl.com/0035.搜索插入位置.html)和[数组：这个循环可以转懵很多人！](https://programmercarl.com/0059.螺旋矩阵II.html)中都强调过循环不变量的重要性，在二分查找以及螺旋矩阵的求解中，坚持循环不变量非常重要，本题也是。
 
diff --git a/problems/0860.柠檬水找零.md b/problems/0860.柠檬水找零.md
index 01bd1a3b..f48ecf4d 100644
--- a/problems/0860.柠檬水找零.md
+++ b/problems/0860.柠檬水找零.md
@@ -128,24 +128,24 @@ public:
 ```java
 class Solution {
     public boolean lemonadeChange(int[] bills) {
-        int cash_5 = 0;
-        int cash_10 = 0;
+        int five = 0;
+        int ten = 0;
 
         for (int i = 0; i < bills.length; i++) {
             if (bills[i] == 5) {
-                cash_5++;
+                five++;
             } else if (bills[i] == 10) {
-                cash_5--;
-                cash_10++;
+                five--;
+                ten++;
             } else if (bills[i] == 20) {
-                if (cash_10 > 0) {
-                    cash_10--;
-                    cash_5--;
+                if (ten > 0) {
+                    ten--;
+                    five--;
                 } else {
-                    cash_5 -= 3;
+                    five -= 3;
                 }
             }
-            if (cash_5 < 0 || cash_10 < 0) return false;
+            if (five < 0 || ten < 0) return false;
         }
         
         return true;

From 1a553b01d691478c049fd39e59548f3a71c9f456 Mon Sep 17 00:00:00 2001
From: erdengk <15596570256@163.com>
Date: Wed, 16 Feb 2022 17:43:36 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200491.=E9=80=92?=
 =?UTF-8?q?=E5=A2=9E=E5=AD=90=E5=BA=8F=E5=88=97.md=20Java=E8=A7=A3?=
 =?UTF-8?q?=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0491.递增子序列.md | 34 +++++++++++++++++++++++++++++++-
 1 file changed, 33 insertions(+), 1 deletion(-)

diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md
index 6d88119e..298586c7 100644
--- a/problems/0491.递增子序列.md
+++ b/problems/0491.递增子序列.md
@@ -227,7 +227,39 @@ class Solution {
     }
 }
 ```
-
+```java
+//法二：使用map
+class Solution {
+    //结果集合
+    List<List<Integer>> res = new ArrayList<>();
+    //路径集合
+    LinkedList<Integer> path = new LinkedList<>();
+    public List<List<Integer>> findSubsequences(int[] nums) {
+        getSubsequences(nums,0);
+        return res;
+    }
+    private void getSubsequences( int[] nums, int start ) {
+        if(path.size()>1 ){
+            res.add( new ArrayList<>(path) );
+            // 注意这里不要加return，要取树上的节点
+        }
+        HashMap<Integer,Integer> map = new HashMap<>();
+        for(int i=start ;i < nums.length ;i++){
+            if(!path.isEmpty() && nums[i]< path.getLast()){
+                continue;
+            }
+            // 使用过了当前数字
+            if ( map.getOrDefault( nums[i],0 ) >=1 ){
+                continue;
+            }
+            map.put(nums[i],map.getOrDefault( nums[i],0 )+1);
+            path.add( nums[i] );
+            getSubsequences( nums,i+1 );
+            path.removeLast();
+        }
+    }
+}
+```
 
 ### Python