From b29529120be5d646700773824c43e9ace450dfd9 Mon Sep 17 00:00:00 2001
From: Andy <2311566266@qq.com>
Date: Mon, 26 Jul 2021 10:10:44 +0800
Subject: [PATCH 1/3] =?UTF-8?q?Update=200225.=E7=94=A8=E9=98=9F=E5=88=97?=
 =?UTF-8?q?=E5=AE=9E=E7=8E=B0=E6=A0=88.md?=
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Content-Transfer-Encoding: 8bit

fix typo.
---
 problems/0225.用队列实现栈.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md
index 85b981e5..03c24132 100644
--- a/problems/0225.用队列实现栈.md
+++ b/problems/0225.用队列实现栈.md
@@ -38,7 +38,7 @@ https://leetcode-cn.com/problems/implement-stack-using-queues/
 
 **队列模拟栈，其实一个队列就够了**，那么我们先说一说两个队列来实现栈的思路。
 
-**队列是先进先出的规则，把一个队列中的数据导入另一个队列中，数据的顺序并没有变，并有变成先进后出的顺序。**
+**队列是先进先出的规则，把一个队列中的数据导入另一个队列中，数据的顺序并没有变，并没有变成先进后出的顺序。**
 
 所以用栈实现队列， 和用队列实现栈的思路还是不一样的，这取决于这两个数据结构的性质。
 

From c08686cc6b26fdd92bb4726080edd0a4ee1e434e Mon Sep 17 00:00:00 2001
From: X-shuffle <53906918+X-shuffle@users.noreply.github.com>
Date: Tue, 27 Jul 2021 10:48:11 +0800
Subject: [PATCH 2/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200135.=E5=88=86?=
 =?UTF-8?q?=E5=8F=91=E7=B3=96=E6=9E=9C=20GO=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
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添加 0135.分发糖果 GO版本
---
 problems/0135.分发糖果.md | 38 ++++++++++++++++++++++++++++++++++-
 1 file changed, 37 insertions(+), 1 deletion(-)

diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md
index fd791277..2d3fca84 100644
--- a/problems/0135.分发糖果.md
+++ b/problems/0135.分发糖果.md
@@ -175,7 +175,43 @@ class Solution:
 ```
 
 Go：
-
+```golang
+func candy(ratings []int) int {
+    /**先确定一边，再确定另外一边
+        1.先从左到右，当右边的大于左边的就加1
+        2.再从右到左，当左边的大于右边的就再加1
+    **/
+    need:=make([]int,len(ratings))
+    sum:=0
+    //初始化(每个人至少一个糖果)
+     for i:=0;i<len(ratings);i++{
+         need[i]=1
+     }
+     //1.先从左到右，当右边的大于左边的就加1
+    for i:=0;i<len(ratings)-1;i++{
+        if ratings[i]<ratings[i+1]{
+            need[i+1]=need[i]+1
+        }
+    }
+    //2.再从右到左，当左边的大于右边的就右边加1，但要花费糖果最少，所以需要做下判断
+    for i:=len(ratings)-1;i>0;i--{
+        if ratings[i-1]>ratings[i]{
+            need[i-1]=findMax(need[i-1],need[i]+1)
+        }
+    }
+    //计算总共糖果
+    for i:=0;i<len(ratings);i++{
+        sum+=need[i]
+    }
+    return sum
+}
+func findMax(num1 int ,num2 int) int{
+    if num1>num2{
+        return num1
+    }
+    return num2
+}
+```
 Javascript:
 ```Javascript
 var candy = function(ratings) {

From 600f82d9166a5b1a3adb3fa84a4421a979fafec7 Mon Sep 17 00:00:00 2001
From: X-shuffle <53906918+X-shuffle@users.noreply.github.com>
Date: Tue, 27 Jul 2021 13:07:18 +0800
Subject: [PATCH 3/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200860.=E6=9F=A0?=
 =?UTF-8?q?=E6=AA=AC=E6=B0=B4=E6=89=BE=E9=9B=B6=20GO=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
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添加 0860.柠檬水找零 GO版本
---
 problems/0860.柠檬水找零.md | 39 ++++++++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)

diff --git a/problems/0860.柠檬水找零.md b/problems/0860.柠檬水找零.md
index a18c008d..0d5a7075 100644
--- a/problems/0860.柠檬水找零.md
+++ b/problems/0860.柠檬水找零.md
@@ -183,6 +183,45 @@ class Solution:
 
 Go：
 
+```golang
+func lemonadeChange(bills []int) bool {
+    //left表示还剩多少 下表0位5元的个数 ，下表1为10元的个数
+    left:=[2]int{0,0}
+    //第一个元素不为5，直接退出
+    if bills[0]!=5{
+        return false
+    }
+    for i:=0;i<len(bills);i++{
+        //先统计5元和10元的个数
+        if bills[i]==5{
+            left[0]+=1
+        }
+        if bills[i]==10{
+            left[1]+=1
+        }
+        //接着处理找零的
+        tmp:=bills[i]-5
+        if tmp==5{
+            if left[0]>0{
+                left[0]-=1
+            }else {
+                return false
+            }
+        }
+        if tmp==15{
+            if left[1]>0&&left[0]>0{
+                left[0]-=1
+                left[1]-=1
+            }else if left[1]==0&&left[0]>2{
+                left[0]-=3
+            }else{
+                return false
+            }
+        }
+    }
+    return true
+}
+```
 
 Javascript:
 ```Javascript