拼多多笔试

code/src/main/java/com/raorao/interview/pdd/Carts.java

@@ -0,0 +1,16 @@
+package com.raorao.interview.pdd;
+
+/**
+ * 去卡片.
+ *
+ * 题目描述：
+ *
+ * @author Xiong Raorao
+ * @since 2018-08-05-20:40
+ */
+public class Carts {
+
+  public static void main(String[] args) {
+
+  }
+}

code/src/main/java/com/raorao/interview/pdd/Friends.java

@@ -0,0 +1,75 @@
+package com.raorao.interview.pdd;
+
+import java.util.ArrayList;
+import java.util.Scanner;
+
+/**
+ * .
+ *
+ * @author Xiong Raorao
+ * @since 2018-08-05-20:03
+ */
+public class Friends {
+
+  public static void main(String[] args) {
+    Scanner scanner = new Scanner(System.in);
+    while (scanner.hasNextLine()) {
+      String[] input = scanner.nextLine().split(" ");
+      int userNum = Integer.parseInt(input[0]);
+      int targetUser = Integer.parseInt(input[1]);
+      int[][] relation = new int[userNum][userNum];
+      for (int i = 0; i < userNum && scanner.hasNextLine(); i++) {
+        String ss = scanner.nextLine();
+        String[] users = ss.split(" ");
+        for (String s : users) {
+          int user = Integer.parseInt(s);
+          relation[i][user] = 1;
+          relation[i][i] = 1;
+        }
+        //System.out.println(" line " + i );
+      }
+      //System.out.println(" out: ");
+      System.out.println(process(relation, targetUser, userNum));
+    }
+
+  }
+
+  private static int process(int[][] relation, int targetUser, int userNum) {
+    // 判断最可能的好友
+    ArrayList<Integer> stranger = new ArrayList<>();
+    for (int i = 0; i < userNum; i++) {
+      if (relation[targetUser][i] != 1) {
+        stranger.add(i);
+      }
+    }
+
+    if (userNum == 1) {
+      return -1;
+    }
+
+    // 判断哪个陌生人和自己的共同好友多
+    if (stranger.size() == 1) {
+      return stranger.get(0);
+    }
+    int resultUser = 0;
+    int maxFriends = getSameFriend(relation, stranger.get(0), targetUser, userNum);
+    for (int i = 1; i < stranger.size(); i++) {
+      int temp = getSameFriend(relation, stranger.get(i), targetUser, userNum);
+      if (temp > maxFriends) {
+        maxFriends = temp;
+        resultUser = stranger.get(i);
+      }
+    }
+    return resultUser;
+  }
+
+  private static int getSameFriend(int[][] relation, int user1, int user2, int userNum) {
+    int count = 0;
+    for (int i = 0; i < userNum; i++) {
+      if (relation[user1][i] == 1 && relation[user2][i] == 1) {
+        count++;
+      }
+    }
+    return count;
+  }
+}

code/src/main/java/com/raorao/interview/pdd/ListComplete.java

@@ -0,0 +1,90 @@
+package com.raorao.interview.pdd;
+
+import java.util.Scanner;
+
+/**
+ * 列表补全问题.
+ *
+ * 描述： 在商城的某个位置有一个商品列表，该列表是由L1、L2两个子列表拼接而成。当用户浏览并翻页时，需要从列表L1、L2中获取商品进行展示。展示规则如下：
+ *
+ * 1. 用户可以进行多次翻页，用offset表示用户在之前页面已经浏览的商品数量，比如offset为4，表示用户已经看了4个商品
+ *
+ * 2. n表示当前页面需要展示的商品数量
+ *
+ * 3. 展示商品时首先使用列表L1，如果列表L1长度不够，再从列表L2中选取商品
+ *
+ * 4. 从列表L2中补全商品时，也可能存在数量不足的情况
+ *
+ * 请根据上述规则，计算列表L1和L2中哪些商品在当前页面被展示了
+ *
+ * 输入描述：
+ *
+ * 每个测试输入包含1个测试用例，包含四个整数，分别表示偏移量offset、元素数量n，列表L1的长度l1，列表L2的长度l2。
+ *
+ * 输出描述：
+ *
+ * 一行内输出四个整数分别表示L1和L2的区间start1，end1，start2，end2，每个数字之间有一个空格。 注意，区间段使用半开半闭区间表示，即包含起点，不包含终点。如果某个列表的区间为空，使用[0,
+ * 0)表示，如果某个列表被跳过，使用[len, len)表示，len表示列表的长度。
+ *
+ * @author Xiong Raorao
+ * @since 2018-08-05-16:02
+ */
+public class ListComplete {
+
+  public static String process(String list) {
+    String[] arr = list.split(" ");
+    int offset = Integer.parseInt(arr[0]);
+    int n = Integer.parseInt(arr[1]);
+    int L1 = Integer.parseInt(arr[2]);
+    int L2 = Integer.parseInt(arr[3]);
+
+    int[] out = new int[4];
+
+    if (offset + n < L1) {
+      out[0] = offset;
+      out[1] = offset + n;
+      out[2] = 0;
+      out[3] = 0;
+    } else if (offset < L1 && offset + n >= L1) {
+      out[0] = offset;
+      out[1] = L1;
+      out[2] = 0;
+      if (offset + n < L1 + L2) {
+        out[3] = offset + n - L1;
+      } else {
+        out[3] = L2;
+      }
+    } else if (offset >= L1 && offset <= L1 + L2) {
+      out[0] = L1;
+      out[1] = L1;
+      out[2] = offset - L1;
+      if (offset + n < L1 + L2) {
+        out[3] = offset + n - L1;
+      } else {
+        out[3] = L2;
+      }
+    } else if (offset > L1 + L2){
+      out[0] = L1;
+      out[1] = L1;
+      out[2] = L2;
+      out[3] = L2;
+    }
+
+    StringBuilder sb = new StringBuilder();
+    sb.append(out[0]).append(" ")
+        .append(out[1]).append(" ")
+        .append(out[2]).append(" ")
+        .append(out[3]);
+    return sb.toString();
+
+  }
+
+
+  public static void main(String[] args) {
+    Scanner scanner = new Scanner(System.in);
+    while (scanner.hasNextLine()) {
+      String line = scanner.nextLine();
+      System.out.println(process(line));
+    }
+  }
+}

code/src/main/java/com/raorao/interview/pdd/MaxMultiply.java

@@ -0,0 +1,35 @@
+package com.raorao.interview.pdd;
+
+import java.util.Scanner;
+
+/**
+ * 最大乘积.
+ *
+ * @author Xiong Raorao
+ * @since 2018-08-05-17:03
+ */
+public class MaxMultiply {
+
+  private static int[] arr;
+  private static int size;
+
+  public static void main(String[] args) {
+    for (int i = 0; i < size; i++) {
+
+    }
+  }
+
+  public static void read() {
+    Scanner scanner = new Scanner(System.in);
+    if (scanner.hasNextInt()) {
+      size = scanner.nextInt();
+      arr = new int[size];
+    }
+    if (scanner.hasNextLine()) {
+      String[] ss = scanner.nextLine().split(" ");
+      for (int i = 0; i < ss.length; i++) {
+        arr[i] = Integer.parseInt(ss[i]);
+      }
+    }
+  }
+}

code/src/main/java/com/raorao/interview/pdd/Numbers.java

@@ -0,0 +1,58 @@
+package com.raorao.interview.pdd;
+
+import java.util.Scanner;
+
+/**
+ * 有趣的变换.
+ *
+ * @author Xiong Raorao
+ * @since 2018-08-05-19:52
+ */
+public class Numbers {
+
+  public static int intSum(String string) {
+    if (string.equals("0")) {
+      return 1;
+    }
+    if (string.startsWith("0")) {
+      return 0;
+    } else {
+      return 1;
+    }
+  }
+
+  public static int foatSum(String string) {
+    if (string.endsWith("0")) {
+      return 0;
+    }
+    if (string.startsWith("0")) {
+      return 1;
+    } else {
+      return string.length() - 1;
+    }
+
+
+  }
+
+  public static int process(String string) {
+    int count = 0;
+    for (int i = 1; i < string.length(); i++) {
+      count +=
+          intSum(string.substring(0, i)) * foatSum(string.substring(i, string.length())) +
+              intSum(string.substring(0, i)) * intSum(string.substring(i, string.length()))
+              +
+              foatSum(string.substring(0, i)) * intSum(
+                  string.substring(i, string.length())) +
+              foatSum(string.substring(0, i)) * foatSum(
+                  string.substring(i, string.length()));
+    }
+    return count;
+  }
+
+  public static void main(String[] args) {
+    Scanner scanner = new Scanner(System.in);
+    String input = scanner.nextLine();
+    System.out.println(process(input));
+  }
+
+}

code/src/main/java/com/raorao/interview/pdd/Square.java

@@ -0,0 +1,72 @@
+package com.raorao.interview.pdd;
+
+import java.util.Scanner;
+
+/**
+ * 旋转的字符串.
+ *
+ * 题目要求： 求给定的数字的字符串
+ *
+ * @author Xiong Raorao
+ * @since 2018-08-05-19:04
+ */
+public class Square {
+
+  private static void process(String ss) {
+    char[] arr = ss.toCharArray();
+    int a;
+    char[][] result;
+    if (ss.length() == 0) {
+      System.out.println();
+    }
+    if (ss.length() % 4 == 0) {
+      a = ss.length() / 4 + 1;
+      result = new char[a][a];
+
+      // 第一行
+      for (int i = 0; i < a; i++) {
+        result[0][i] = arr[i];
+      }
+
+      // 最后一行
+      for (int i = 0; i < a; i++) {
+        result[a - 1][a - i - 1] = arr[2 * a - 2 + i];
+      }
+
+      // 右边
+      for (int i = 0; i < a - 2; i++) {
+        result[i + 1][a - 1] = arr[a + i];
+      }
+
+      // 左边
+      for (int i = 0; i < a - 2; i++) {
+        result[a - 2 - i][0] = arr[3 * a - 2 + i];
+      }
+
+      // 打印
+      print(result);
+    }
+  }
+
+  public static void main(String[] args) {
+    Scanner scanner = new Scanner(System.in);
+    while (scanner.hasNextLine()) {
+      process(scanner.nextLine());
+    }
+  }
+
+  private static void print(char[][] arr) {
+    int a = arr.length;
+    for (int i = 0; i < a; i++) {
+      for (int j = 0; j < a; j++) {
+        if (i == 0 || j == 0 || i == a - 1 || j == a - 1) {
+          System.out.print(arr[i][j] + "");
+        } else {
+          System.out.print(" ");
+        }
+      }
+      System.out.println();
+    }
+  }
+
+}

interview/README.md

@@ -66,6 +66,12 @@ df, disk free, 通过文件系统来快速获取空间大小的信息，当我
 (6) TCP 四次挥手的过程？time_out的缺点？  
 (7) mysql 的 group by 和 partition by?  
 答：group by 用来做聚合操作，partition by 用于对某个字段分区，然后做某些操作。  
+
+普通的聚合函数用group by分组，每个分组返回一个统计值，而分析函数采用partition by分组，并且每组每行都可以返回一个统计值。
+分析函数的形式：分析函数带有一个开窗函数over()，包含三个分析子句:分组(partition by), 排序(order by), 窗口(rows) 。
+
+parition by 针对Oracle数据库
+
 [参考链接](https://blog.csdn.net/cyl937/article/details/19930349)
 (8) 海量数据去重，例如大量的IP地址，如何去重？  
 答：MapReduce