leecode更新

Leecode/src/main/java/com/markilue/leecode/backtrace/T01_Combine.java

@@ -16,7 +16,7 @@ import java.util.List;
  *  你可以按 任何顺序 返回答案。
  * @Version: 1.0
  */
-public class Combine {
+public class T01_Combine {
 
 
     @Test

Leecode/src/main/java/com/markilue/leecode/backtrace/T02_CombinationSum3.java

@@ -17,7 +17,7 @@ import java.util.List;
  * 返回 所有可能的有效组合的列表 。该列表不能包含相同的组合两次，组合可以以任何顺序返回。
  * @Version: 1.0
  */
-public class combinationSum3 {
+public class T02_CombinationSum3 {
 
 
     @Test

Leecode/src/main/java/com/markilue/leecode/backtrace/T03_LetterCombinations.java

@@ -14,7 +14,7 @@ import java.util.*;
  * 给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
  * @Version: 1.0
  */
-public class IetterCombinations {
+public class T03_LetterCombinations {
 
     @Test
     public void test() {

Leecode/src/main/java/com/markilue/leecode/backtrace/second/T01_Combine.java

@@ -0,0 +1,58 @@
+package com.markilue.leecode.backtrace.second;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-31  10:41
+ *@Description:
+ * TODO  二刷力扣77题  组合:
+ *  给定两个整数 n 和 k，返回范围 [1, n] 中所有可能的 k 个数的组合。
+ *  你可以按 任何顺序 返回答案。
+ *@Version: 1.0
+ */
+public class T01_Combine {
+
+    @Test
+    public void test() {
+        int n = 4;
+        int k = 2;
+        System.out.println(combine(n, k));
+    }
+
+    List<List<Integer>> result = new ArrayList<>();
+    List<Integer> cur = new ArrayList<>();
+
+    public List<List<Integer>> combine(int n, int k) {
+        backtracking(n, k, 1);
+        return result;
+
+    }
+
+    /**
+     * 回溯法：
+     * 剪枝前速度击败35.29%，内存击败55.4%  17ms
+     * 剪枝后速度击败99.99%，内存击败74.46%  1ms
+     * @param n
+     * @param k
+     * @param now
+     */
+    public void backtracking(int n, int k, int now) {
+        if (cur.size() == k) {
+            List<Integer> list = new ArrayList<>(cur);
+            result.add(list);
+            return;
+        }
+        for (int i = now; i <= n-(k-cur.size())+1; i++) {
+            //剪枝，后面怎么都不够了肯定不用了
+            cur.add(i);
+            backtracking(n, k, i + 1);
+            cur.remove(cur.size() - 1);
+        }
+    }
+}

Leecode/src/main/java/com/markilue/leecode/backtrace/second/T02_CombinationSum3.java

@@ -0,0 +1,61 @@
+package com.markilue.leecode.backtrace.second;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-31  11:07
+ *@Description:
+ * TODO  二刷力扣216题  组合总和III:
+ *  找出所有相加之和为 n 的 k 个数的组合，且满足下列条件：
+ *    只使用数字1到9
+ *    每个数字 最多使用一次
+ *  返回 所有可能的有效组合的列表 。该列表不能包含相同的组合两次，组合可以以任何顺序返回。
+ *@Version: 1.0
+ */
+public class T02_CombinationSum3 {
+
+    @Test
+    public void test() {
+        int k = 3, n = 7;
+        System.out.println(combinationSum3(k, n));
+    }
+
+    @Test
+    public void test1() {
+        int k = 9, n = 45;
+        System.out.println(combinationSum3(k, n));
+    }
+
+
+    List<List<Integer>> result = new ArrayList<>();
+    List<Integer> cur = new ArrayList<>();
+
+    public List<List<Integer>> combinationSum3(int k, int n) {
+        backtracking(k, n, 0, 1);
+        return result;
+    }
+
+    public void backtracking(int k, int n, int sum, int now) {
+        if (k == cur.size()) {
+            if (sum == n) {
+                result.add(new ArrayList<>(cur));
+            }
+            return;
+        }
+        int flag = n - sum > 9 ? 9 : n - sum;
+        for (int i = now; i <= flag - (k - cur.size()) + 1; i++) {
+            cur.add(i);
+            sum += i;
+            backtracking(k, n, sum, i + 1);
+            sum -= i;
+            cur.remove(cur.size() - 1);
+        }
+
+    }
+}

Leecode/src/main/java/com/markilue/leecode/backtrace/second/T03_LetterCombinations.java

@@ -0,0 +1,70 @@
+package com.markilue.leecode.backtrace.second;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-31  11:34
+ *@Description:
+ * TODO  二刷力扣17题  电话号码的字母组合:
+ *  给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。
+ *  给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
+ *@Version: 1.0
+ */
+public class T03_LetterCombinations {
+
+    @Test
+    public void test() {
+//        String s = "01";
+//        for (int i = 0; i < s.length(); i++) {
+//            System.out.println(s.charAt(i) - '0');
+//        }
+        String digits = "23";
+        System.out.println(letterCombinations(digits));
+
+    }
+
+    List<String> result = new ArrayList<>();
+    StringBuilder builder = new StringBuilder();
+    Map<Character, char[]> map = new HashMap<Character, char[]>() {{
+        put('2', new char[]{'a', 'b', 'c'});
+        put('3', new char[]{'d', 'e', 'f'});
+        put('4', new char[]{'g', 'h', 'i'});
+        put('5', new char[]{'j', 'k', 'l'});
+        put('6', new char[]{'m', 'n', 'o'});
+        put('7', new char[]{'p', 'q', 'r', 's'});
+        put('8', new char[]{'t', 'u', 'v'});
+        put('9', new char[]{'w', 'x', 'y', 'z'});
+    }};
+
+    public List<String> letterCombinations(String digits) {
+        if(digits.length()==0){
+            return result;
+        }
+
+        char[] chars = digits.toCharArray();
+        backtracking(chars, chars.length, 0);
+
+        return result;
+    }
+
+    public void backtracking(char[] chars, int total, int now) {
+        if (builder.length() == total) {
+            result.add(builder.toString());
+            return;
+        }
+        for (char c : map.get(chars[now])) {
+            builder.append(c);
+            backtracking(chars, total, now + 1);
+            builder.deleteCharAt(builder.length() - 1);
+        }
+
+    }
+}

Leecode/src/main/java/com/markilue/leecode/tree/T14_1_BuildTreeII.java

@@ -18,7 +18,7 @@ public class T14_1_BuildTreeII {
     @Test
     public void test() {
         int[] preorder = {3, 9, 20, 15, 7}, inorder = {9, 3, 15, 20, 7};
-        TreeNode node = buildTree(preorder, inorder);
+        TreeNode node = buildTree1(preorder, inorder);
 
         System.out.println(levelOrder(node));
     }

TensorFlow_eaxmple/Model_train_test/spider/qidian.py

@@ -0,0 +1,21 @@
+import scrapy
+import requests
+
+
+class qidianSpider(scrapy.Spider):
+    name = 'qidian'
+
+    start_urls =['https://vipreader.qidian.com/chapter/1031940621/742792391']
+
+    def parse(self, response, **kwargs):
+        divs=response.xpath('//div[@class="read-content j_readContent"]')
+        print(divs)
+
+
+
+if __name__ == '__main__':
+    url = "https://vipreader.qidian.com/chapter/1031940621/742792391/"
+
+    response = requests.get(url=url)
+
+    qidianSpider().parse(response=response)

TensorFlow_eaxmple/Model_train_test/spider/test.py

@@ -0,0 +1,23 @@
+import  requests
+from lxml import etree
+import scrapy
+
+
+url="https://vipreader.qidian.com/chapter/1031940621/742792391/"
+
+response=requests.get(url=url)
+# print(response.text)
+
+html=etree.HTML(response.text)
+total=html.xpath('//*[@id="j_742792391"]/p')
+# total=html.xpath('//*[@id="j_742792391"]/p[15]/text()')
+total=html.xpath('//div[@class="read-content j_readContent"]/p')
+print(total)
+textList=[]
+for i in range(10):
+    # print(p)
+    textList.append(html.xpath('//*[@id="j_742792391"]/p[{0}]/text()'.format(i)))
+    # textList.append(html.xpath('//*[@id="j_742792391"]/p[1]/text()'))
+# print(text1)
+# print(text2)
+print(textList)