diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/IetterCombinations.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/IetterCombinations.java
index 47b17d2..8a85d1e 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/IetterCombinations.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/IetterCombinations.java
@@ -1,6 +1,5 @@
 package com.markilue.leecode.backtrace;
 
-import com.markilue.leecode.stackAndDeque.EvalRPN;
 import org.junit.Test;
 
 import java.util.*;
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java
index c41236b..bea6c89 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java
@@ -1,9 +1,7 @@
 package com.markilue.leecode.backtrace;
 
-import com.markilue.leecode.stackAndDeque.EvalRPN;
 import org.junit.Test;
 
-import javax.print.DocFlavor;
 import java.util.*;
 
 /**
diff --git a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T04_RemoveDuplicates.java b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T04_RemoveDuplicates.java
new file mode 100644
index 0000000..4089d71
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T04_RemoveDuplicates.java
@@ -0,0 +1,105 @@
+package com.markilue.leecode.stackAndDeque;
+
+import org.junit.Test;
+
+import java.util.Deque;
+import java.util.LinkedList;
+import java.util.Stack;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.stackAndDeque
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-03  10:24
+ *@Description:
+ * TODO  力扣104题  删除组字符串中的所有相邻重复项
+ *  给出由小写字母组成的字符串 S，重复项删除操作会选择两个相邻且相同的字母，并删除它们。
+ *  在 S 上反复执行重复项删除操作，直到无法继续删除。
+ *  在完成所有重复项删除操作后返回最终的字符串。答案保证唯一。
+ *@Version: 1.0
+ */
+public class T04_RemoveDuplicates {
+
+    @Test
+    public void test(){
+        String s= "abbaca";
+        System.out.println(removeDuplicates(s));
+    }
+
+
+    /**
+     * 思路:类似消消乐,可以使用一个栈装，如果和栈顶一样就消栈，否则入栈
+     * 速度击败56.41%，内存击败62.57%  44ms
+     * @param s
+     * @return
+     */
+    public String removeDuplicates(String s) {
+
+        Deque<Character> stack = new LinkedList<Character>();
+        char[] chars = s.toCharArray();
+
+        for (char aChar : chars) {
+            if (!stack.isEmpty() && stack.peek() == aChar) {
+                stack.pop();
+            } else {
+                stack.push(aChar);
+            }
+        }
+        StringBuilder builder =new StringBuilder();
+        //stack全pop出,组成字符串
+        while (!stack.isEmpty()){
+            builder.insert(0,stack.pop());
+        }
+
+        return builder.toString();
+    }
+
+
+    /**
+     * 代码随想录拿字符串做栈
+     * 速度击败71.16%，内存击败96.29%  22ms
+     * @param s
+     * @return
+     */
+    public String removeDuplicates1(String s) {
+        // 将 res 当做栈
+        // 也可以用 StringBuilder 来修改字符串，速度更快
+        // StringBuilder res = new StringBuilder();
+        StringBuffer res = new StringBuffer();
+        // top为 res 的长度
+        int top = -1;
+        for (int i = 0; i < s.length(); i++) {
+            char c = s.charAt(i);
+            // 当 top > 0,即栈中有字符时，当前字符如果和栈中字符相等，弹出栈顶字符，同时 top--
+            if (top >= 0 && res.charAt(top) == c) {
+                res.deleteCharAt(top);
+                top--;
+                // 否则，将该字符 入栈，同时top++
+            } else {
+                res.append(c);
+                top++;
+            }
+        }
+        return res.toString();
+    }
+
+
+    /**
+     * 评论区方法:使用相同了就覆盖,类似于快慢指针
+     * 速度击败96.3%，内存击败74.66%  4ms
+     * @param S
+     * @return
+     */
+    public String removeDuplicates2(String S) {
+        char[] s = S.toCharArray();
+        int top = -1;
+        for (int i = 0; i < S.length(); i++) {
+            if (top == -1 || s[top] != s[i]) {
+                s[++top] = s[i];
+            } else {
+                top--;
+            }
+        }
+        return String.valueOf(s, 0, top + 1);
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/EvalRPN.java b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T05_EvalRPN.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/stackAndDeque/EvalRPN.java
rename to Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T05_EvalRPN.java
index 3a176b3..9a766b8 100644
--- a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/EvalRPN.java
+++ b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T05_EvalRPN.java
@@ -17,7 +17,7 @@ import java.util.*;
  *      可以保证给定的逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
  * @Version: 1.0
  */
-public class EvalRPN {
+public class T05_EvalRPN {
 
     @Test
     public void test(){
diff --git a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/MaxSlidingWindow.java b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T06_MaxSlidingWindow.java
similarity index 97%
rename from Leecode/src/main/java/com/markilue/leecode/stackAndDeque/MaxSlidingWindow.java
rename to Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T06_MaxSlidingWindow.java
index 0573eae..6fd7fcf 100644
--- a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/MaxSlidingWindow.java
+++ b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/T06_MaxSlidingWindow.java
@@ -9,12 +9,14 @@ import java.util.*;
  * @BelongsPackage: com.markilue.leecode.stackAndDeque
  * @Author: dingjiawen
  * @CreateTime: 2022-09-14  09:31
- * @Description: TODO 力扣239题  滑动窗口最大值:
- * 给你一个整数数组 nums，有一个大小为k的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k个数字。滑动窗口每次只向右移动一位。
- * 返回 滑动窗口中的最大值 。
+ * @Description:
+ * TODO 力扣239题  滑动窗口最大值:
+ *  给你一个整数数组 nums，有一个大小为k的滑动窗口从数组的最左侧移动到数组的最右侧。
+ *  你只可以看到在滑动窗口内的 k个数字。滑动窗口每次只向右移动一位。
+ *  返回 滑动窗口中的最大值 。
  * @Version: 1.0
  */
-public class MaxSlidingWindow {
+public class T06_MaxSlidingWindow {
 
 
     @Test
diff --git a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/second/T05_EvalRPN.java b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/second/T05_EvalRPN.java
new file mode 100644
index 0000000..4127a51
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/second/T05_EvalRPN.java
@@ -0,0 +1,162 @@
+package com.markilue.leecode.stackAndDeque.second;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.Stack;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.stackAndDeque.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-03  11:27
+ *@Description:
+ * TODO  力扣150题  逆波兰表达式求值:
+ *  根据 逆波兰表示法，求表达式的值。
+ *      有效的算符包括+、-、*、/。每个运算对象可以是整数，也可以是另一个逆波兰表达式。
+ *      注意两个整数之间的除法只保留整数部分。
+ *      可以保证给定的逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
+ *@Version: 1.0
+ */
+public class T05_EvalRPN {
+
+    @Test
+    public void test(){
+       String[] tokens = {"2", "1", "+", "3", "*"};
+        System.out.println(evalRPN(tokens));
+    }
+
+    @Test
+    public void test1(){
+        String[] tokens = {"4","13","5","/","+"};
+        System.out.println(evalRPN2(tokens));
+    }
+
+
+    /**
+     * 思路：用stack装
+     * 速度击败54.45%，内存击败10.19%  6ms
+     * @param tokens
+     * @return
+     */
+    public int evalRPN(String[] tokens) {
+
+        Stack<Integer> stack = new Stack<>();//存数
+
+        for (String token : tokens) {
+
+            if(token.equals("+")){
+                Integer num2 = stack.pop();
+                Integer num1 = stack.pop();
+                int value=num1+num2;
+                stack.push(value);
+            }else if(token.equals("-")){
+                Integer num2 = stack.pop();
+                Integer num1 = stack.pop();
+                int value=num1-num2;
+                stack.push(value);
+            }else if(token.equals("*")){
+                Integer num2 = stack.pop();
+                Integer num1 = stack.pop();
+                int value=num1*num2;
+                stack.push(value);
+            }else if(token.equals("/")){
+                Integer num2 = stack.pop();
+                Integer num1 = stack.pop();
+                int value=num1/num2;
+                stack.push(value);
+            }else {
+                stack.push(Integer.valueOf(token));
+            }
+        }
+
+        return stack.peek();
+
+    }
+
+
+    /**
+     * 官方的规范写法
+     * @param tokens
+     * @return
+     */
+    public int evalRPN1(String[] tokens) {
+
+        Stack<Integer> stack = new Stack<>();
+
+        ArrayList<String> list = new ArrayList<>(Arrays.asList("+", "-", "*", "/"));
+
+        for (String token : tokens) {
+            if(!list.contains(token)){
+                stack.push(Integer.valueOf(token));
+            }else {
+                int num2=stack.pop();
+                int num1=stack.pop();
+                if(token.equals("+")){
+                    stack.push(num1+num2);
+                }else if(token.equals("-")){
+                    stack.push(num1-num2);
+                }else if(token.equals("*")){
+                    stack.push(num1*num2);
+                }else if(token.equals("/")){
+                    stack.push(num1/num2);
+                }
+            }
+        }
+        return stack.pop();
+
+    }
+
+
+    /**
+     * 官方最快：
+     * 速度击败100%，内存击败31.2%  1ms
+     */
+    int index;
+    String[] tokens;
+
+    public int evalRPN2(String[] tokens) {
+        this.tokens = tokens;
+        index = tokens.length - 1;
+        return recursion();
+    }
+    //递归模拟栈
+    public int recursion() {
+        String sub = tokens[index];
+        switch (sub) {
+            case "+" : {
+                index--;
+                int right = recursion();
+                index--;
+                int left = recursion();
+                return left + right;
+            }
+            case "-" : {
+                index--;
+                int right = recursion();
+                index--;
+                int left = recursion();
+                return left - right;
+            }
+            case "*" : {
+                index--;
+                int right = recursion();
+                index--;
+                int left = recursion();
+                return left * right;
+            }
+            case "/" : {
+                index--;
+                int right = recursion();
+                index--;
+                int left = recursion();
+                return left / right;
+            }
+            default : {
+                return Integer.parseInt(sub);
+            }
+        }
+    }
+
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/second/T06_MaxSlidingWindow.java b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/second/T06_MaxSlidingWindow.java
new file mode 100644
index 0000000..affa70b
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/stackAndDeque/second/T06_MaxSlidingWindow.java
@@ -0,0 +1,156 @@
+package com.markilue.leecode.stackAndDeque.second;
+
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.stackAndDeque.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-03  12:00
+ *@Description:
+ * TODO 力扣239题  滑动窗口最大值:
+ *  给你一个整数数组 nums，有一个大小为k的滑动窗口从数组的最左侧移动到数组的最右侧。
+ *  你只可以看到在滑动窗口内的 k个数字。滑动窗口每次只向右移动一位。
+ *  返回 滑动窗口中的最大值 。
+ *@Version: 1.0
+ */
+public class T06_MaxSlidingWindow {
+
+    @Test
+    public void test() {
+        int[] nums = {1, 3, -1, -3, 5, 3, 6, 7};
+        int k = 3;
+        System.out.println(Arrays.toString(maxSlidingWindow(nums, k)));
+    }
+
+    @Test
+    public void test1() {
+        int[] nums = {1, -1};
+        int k = 1;
+        System.out.println(Arrays.toString(maxSlidingWindow(nums, k)));
+    }
+
+    /**
+     * 思路:维护一个优先队列(单调的)，每次淘汰过期的元素即可
+     * 速度18.32%，内存击败33.44%
+     * @param nums
+     * @param k
+     * @return
+     */
+    public int[] maxSlidingWindow(int[] nums, int k) {
+
+        int[] result = new int[nums.length - k + 1];
+        PriorityQueue<int[]> deque = new PriorityQueue<>(new Comparator<int[]>() {
+            @Override
+            public int compare(int[] o1, int[] o2) {
+                return o1[0] != o2[0] ? o2[0] - o1[0] : o2[1] - o1[1];
+            }
+        });//<value,index>
+
+
+        //构造第一个窗口
+        int max = nums[0]; //维护窗口最大值
+        int index = -1;
+        for (int i = 0; i < k; i++) {
+            deque.offer(new int[]{nums[i], i});
+            if (max < nums[i]) max = nums[i];
+        }
+        result[++index] = max;
+
+        for (int i = k; i < nums.length; i++) {
+            deque.offer(new int[]{nums[i], i});
+            while (i - k >= deque.peek()[1]) {
+                deque.poll();
+            }
+            result[++index] = deque.peek()[0];
+        }
+        return result;
+
+    }
+
+    /**
+     * 代码随想录思路:维护一个双端队列,然后使其单调，每次加入时淘汰过窗口的和非单调的
+     * 速度击败83.77%，内存击败48.77%
+     * @param nums
+     * @param k
+     * @return
+     */
+    public int[] maxSlidingWindow1(int[] nums, int k) {
+        ArrayDeque<Integer> deque = new ArrayDeque<>();
+        int n = nums.length;
+        int[] res = new int[n - k + 1];
+        int idx = 0;
+        for(int i = 0; i < n; i++) {
+            // 根据题意，i为nums下标，是要在[i - k + 1, i] 中选到最大值，只需要保证两点
+            // 1.队列头结点需要在[i - k + 1, i]范围内，不符合则要弹出
+            while(!deque.isEmpty() && deque.peek() < i - k + 1){
+                deque.poll();
+            }
+            // 2.既然是单调，就要保证每次放进去的数字要比末尾的都大，否则也弹出
+            while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
+                deque.pollLast();
+            }
+
+            deque.offer(i);
+
+            // 因为单调，当i增长到符合第一个k范围的时候，每滑动一步都将队列头节点放入结果就行了
+            if(i >= k - 1){
+                res[idx++] = nums[deque.peek()];
+            }
+        }
+        return res;
+    }
+
+    /**
+     * 官方最快：理论上来说是最基础的方法，只不过可能剪枝效果太好，看起来速度很多
+     * 极端情况下需要一直在窗口中for来寻找最大值，时间复杂度直接O(k*n)
+     * 速度击败100%，内存击败99.59%
+     * @param nums
+     * @param k
+     * @return
+     */
+    public int[] maxSlidingWindow2(int[] nums, int k) {
+        int n = nums.length;
+        int left = 0, right = k - 1;
+        // 当前的最大值的下标
+        int pre = -1;
+        int max = Integer.MIN_VALUE; // min - 1 = max
+        int[] ans = new int[n - k + 1];
+
+        while (right < n) {
+            // 滑动窗口left-right
+            if (left <= pre) {
+                // 窗口滑动前的最大值，还在窗口里面
+                // 比较上一轮最大值和窗口新增的值
+                if (nums[right] >= nums[pre]) {
+                    pre = right;
+                    max = nums[right];
+                }
+            } else if (nums[right] >= max - 1 ) {
+                //max-1是因为max是right-left中最大的，如果有相同的他肯定是最后一个
+                //所以这个数后面的数如果不是最大值，那么肯定<=max-1
+                pre = right;
+                max = nums[right];
+            } else if (nums[left] >= max - 1) {
+                pre = left;
+                max = nums[left];
+            } else {
+                //既不是最左边又不是最右边，就需要手动求最大值
+                max = Integer.MIN_VALUE;
+                // 求当前窗口的最大值
+                for (int i = left; i <= right; i++) {
+                    if (nums[i] >= max) {
+                        max = nums[i];
+                        pre = i;
+                    }
+                }
+            }
+            ans[left] = max;
+            left++;
+            right++;
+        }
+        return ans;
+    }
+}
diff --git a/TensorFlow_eaxmple/Model_train_test/condition_monitoring/test.py b/TensorFlow_eaxmple/Model_train_test/condition_monitoring/test.py
index 8a3d6e5..11e2a91 100644
--- a/TensorFlow_eaxmple/Model_train_test/condition_monitoring/test.py
+++ b/TensorFlow_eaxmple/Model_train_test/condition_monitoring/test.py
@@ -6,7 +6,8 @@ import numpy as np
 import random
 import pandas as pd
 import seaborn as sns
-from condition_monitoring.data_deal.loadData import read_data
+import scipy.signal
+
 
 '''
 @Author : dingjiawen
@@ -193,17 +194,6 @@ def test_mse(mse_file_name:str=mse_file_name,max_file_name:str=max_file_name):
 
 
 
-
-def test_corr(file_name=source_path,N=10):
-    needed_data, label = read_data(file_name=file_name, isNew=False)
-    print(needed_data)
-    print(needed_data.shape)
-    # plot_original_data(needed_data)
-    person = plot_Corr(needed_data, label)
-    person = np.array(person)
-    pass
-
-
 def plot_raw_data():
     # data, label = read_data(file_name='G:\data\SCADA数据\jb4q_8_delete_total_zero.csv', isNew=False)
     # data = np.loadtxt('G:\data\SCADA数据/normalization.csv',delimiter=',')
@@ -218,7 +208,14 @@ if __name__ == '__main__':
     # test_mse()
     # test_result()
     # test_corr()
-    plot_raw_data()
+    # plot_raw_data()
+    data=np.load("H:\data\predict_data\\HI_merge_data.npy")
+    print(data)
+    data=data[:1250,1]
+    plt.plot(data)
+    data_smooth=scipy.signal.savgol_filter(data,53,3)
+    plt.plot(data_smooth)
+    plt.show()
     pass