From f13333e69ba9ee2a9d67392232693a8d18b37692 Mon Sep 17 00:00:00 2001 From: kevinding1125 <745518019@qq.com> Date: Fri, 9 Jun 2023 12:03:15 +0800 Subject: [PATCH] =?UTF-8?q?leecode=EF=BC=8Crt=5Fphm=E6=9B=B4=E6=96=B0?= =?UTF-8?q?=E6=9B=B4=E6=96=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../streaming/SparkStreaming02_Queue.scala | 4 +- .../SparkStreaming06_state_transform.scala | 6 +- .../second/T72_239_MaxSlidingWindow.java | 35 ++++- .../interview/meituan/T0415/Question2.java | 49 ++++++ .../interview/meituan/T0415/Question3.java | 142 ++++++++++++++++++ 5 files changed, 230 insertions(+), 6 deletions(-) create mode 100644 Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question2.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question3.java diff --git a/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming02_Queue.scala b/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming02_Queue.scala index bc0005f..8d04cbd 100644 --- a/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming02_Queue.scala +++ b/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming02_Queue.scala @@ -2,7 +2,7 @@ package com.atguigu.spark.streaming import org.apache.spark.SparkConf import org.apache.spark.rdd.RDD -import org.apache.spark.streaming.dstream.ReceiverInputDStream +import org.apache.spark.streaming.dstream.{InputDStream, ReceiverInputDStream} import org.apache.spark.streaming.{Seconds, StreamingContext} import scala.collection.mutable @@ -27,7 +27,7 @@ object SparkStreaming02_Queue { val rddQueue = new mutable.Queue[RDD[Int]]() //4.创建QueueInputDStream - val inputStream = ssc.queueStream(rddQueue,oneAtATime = false) + val inputStream: InputDStream[Int] = ssc.queueStream(rddQueue,oneAtATime = false) //5.处理队列中的RDD数据 diff --git a/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming06_state_transform.scala b/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming06_state_transform.scala index a23692d..0d0ac2b 100644 --- a/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming06_state_transform.scala +++ b/Big_data_example/Spark/src/main/java/com/atguigu/spark/streaming/SparkStreaming06_state_transform.scala @@ -29,10 +29,10 @@ object SparkStreaming06_state_transform { //和直接map的区别:写code的位置 //Driver端 val newDS: DStream[String] = lines.transform( - rdd =>{ + rdd => { //code:Driver端，(周期性执行) rdd.map( - str =>{ + str => { //code:Executor端 str } @@ -42,7 +42,7 @@ object SparkStreaming06_state_transform { //code:Driver端 val newDS1: DStream[String] = lines.map( - str =>{ + str => { //code:executor端，无周期性执行 str } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T72_239_MaxSlidingWindow.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T72_239_MaxSlidingWindow.java index a879646..668fa54 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T72_239_MaxSlidingWindow.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T72_239_MaxSlidingWindow.java @@ -19,7 +19,7 @@ public class T72_239_MaxSlidingWindow { public void test() { int[] nums = {1, 3, -1, -3, 5, 3, 6, 7}; int k = 3; - System.out.println(Arrays.toString(maxSlidingWindow1(nums, k))); + System.out.println(Arrays.toString(maxSlidingWindow2(nums, k))); } @Test @@ -110,4 +110,37 @@ public class T72_239_MaxSlidingWindow { } + //构造一个单调栈，只有后面的值比前面的值小，前面的值就是无用的 + public int[] maxSlidingWindow2(int[] nums, int k) { + + ArrayDeque stack = new ArrayDeque<>(); + + int[] result = new int[nums.length - k + 1]; + + //构造第一个窗口 + for (int i = 0; i < k; i++) { + while (!stack.isEmpty() && nums[stack.peekLast()] <= nums[i]) { + stack.pollLast(); + } + stack.offerLast(i); + } + result[0] = nums[stack.peekFirst()]; + + for (int i = k; i < nums.length; i++) { + //先排除过期元素 + while (!stack.isEmpty() && i - stack.peekFirst() >= k) { + stack.pollFirst(); + } + //添加在合适的位置 + while (!stack.isEmpty() && nums[stack.peekLast()] <= nums[i]) { + stack.pollLast(); + } + stack.offerLast(i); + result[i - k + 1] = stack.isEmpty() ? -1 : nums[stack.peekFirst()]; + } + return result; + + } + + } diff --git a/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question2.java b/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question2.java new file mode 100644 index 0000000..1619641 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question2.java @@ -0,0 +1,49 @@ +package com.markilue.leecode.interview.meituan.T0415; + +import java.util.Scanner; + + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.interview.meituan.T0415 + *@Author: markilue + *@CreateTime: 2023-06-07 10:28 + *@Description: + * TODO 小美分糖: + * 某一天，小美从商店买了两种糖果，分别买了a个和b个，要分给班上n个小朋友。为了不浪费，每块糖果都得恰好分到一个小朋友。 + * 另外，两种糖果一起吃的话味道其实并不好，所以每一个小朋友都只能得到其中一种糖果。 + * 小美希望分得最少糖果的那个小朋友能得到尽量多的糖果。小美的任务是求得这个数量是多少。 + *@Version: 1.0 + */ +public class Question2 { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + int count = sc.nextInt(); + + for (int i = 0; i < count; i++) { + int n = sc.nextInt(); + int a = sc.nextInt(); + int b = sc.nextInt(); + solve(n, a, b); + } + } + + //挨个遍历均分，看看谁的最小更大 + private static void solve(int n, int a, int b) { + if (a > b) { + solve(n, b, a); + return; + } + int min = Integer.MIN_VALUE; + for (int i = 1; i < n; i++) { + int curMin = Math.min(a / i, b / (n - i)); + if (min > curMin) { + break;//在递减了，直接break + } else { + min = curMin; + } + } + System.out.println(min); + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question3.java b/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question3.java new file mode 100644 index 0000000..98bcd89 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question3.java @@ -0,0 +1,142 @@ +package com.markilue.leecode.interview.meituan.T0415; + +import org.junit.Test; + +import java.util.ArrayList; +import java.util.List; +import java.util.Scanner; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.interview.meituan.T0415 + *@Author: markilue + *@CreateTime: 2023-06-07 11:00 + *@Description: + * TODO 交通规划: + * A国有n个城市，这n个城市排成一列，依次编号为1,2,3,...,n。 + * 一开始，这n座城市之间都没有任何交通路线，于是政府打算修建一些铁路来进行交通规划。 + * 接下来T天，每一天会进行如下操作的其中一种： + * - “L x”：表示编号为 x 的城市与其左边的城市之间修建一条铁路。如果 x 左边没有城市或者已经修建了铁路，则无视该操作； + * - “R x”：表示编号为 x 的城市与其右边的城市之间修建一条铁路。如果 x 右边没有城市或者已经修建了铁路，则无视该操作； + * - “Q x”：表示查询 x 往左边和往右边最远能到达的城市编号。 + * 你的任务是模拟以上操作，并对于每一条“Q x”操作，输出对应的答案。 + *@Version: 1.0 + */ +public class Question3 { + + public static void main(String[] args) { + + Scanner sc = new Scanner(System.in); + int total = sc.nextInt(); + //构建图 + List edge = new ArrayList<>(); + for (int i = 0; i < total; i++) { + edge.add(new Node(i)); + } + + int count = sc.nextInt(); + + for (int i = 0; i < count; i++) { + String action = sc.next(); + int node = sc.nextInt() - 1;//num => index + if (action.equals("L") && node - 1 >= 0) { + edge.get(node).left = edge.get(node - 1); + } else if (action.equals("R") && node + 1 < edge.size()) { + edge.get(node).right = edge.get(node + 1); + } else if (action.equals("Q")) { + //查询 + Node cur = edge.get(node); + Node tempLeft = cur; + while (tempLeft.left != null) { + tempLeft = tempLeft.left; + } + Node tempRight = cur; + while (tempRight.right != null) { + tempRight = tempRight.right; + } + System.out.println((tempLeft.val + 1) + " " + (tempRight.val + 1)); + } + } + + } + + static class Node { + Node left; + Node right; + int val; + + public Node() { + } + + public Node(int val) { + this.val = val; + } + } + + + int[] father; + + public void init(int[] father) { + for (int i = 0; i < father.length; i++) { + father[i] = i; + } + } + + public int find(int x) { + if (x == father[x]) return x; + father[x] = find(father[x]); + return father[x]; + } + + public void union(int u, int v) { + u = find(u); + v = find(v); + if (u == v) return; + father[v] = u; + } + + + //并查集:这题本质上就是一个无向图，考察连通性，因此可以使用并查集简化计算 + public void solve1() { + Scanner sc = new Scanner(System.in); + int total = sc.nextInt(); + //构建图 + father = new int[total + 2]; + init(father); + + + int count = sc.nextInt(); + + for (int i = 0; i < count; i++) { + String action = sc.next(); + int node = sc.nextInt();//num => index + if (action.equals("L")) { + union(node, node - 1); + } else if (action.equals("R")) { + union(node, node + 1); + } else { + //查询 左边二分找到连通的最小值 + int l = 1; + int r = node; + while (l < r) { + int mid = l + ((r - l) >> 1); + if (find(node) == find(mid)) r = mid;//父亲是一样的，我们认为是联通的,所以缩小范围找更小的 + else l = mid + 1; + } + int res1 = r; + //查询 右边二分找到连通的最大值 + l = node; + r = total; + while (l < r) { + int mid = l + ((r - l) >> 1); + if (find(node) == find(mid)) l = mid;//父亲是一样的，我们认为是联通的,所以缩小范围找更大的 + else r = mid - 1; + } + System.out.println(res1 + " " + r); + + } + } + } + + +}