leecode，rt_phm更新更新
This commit is contained in:
kevinding1125
parent
4d08683952
commit
f13333e69b
|
|
@ -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
|
||||
|
|
|
|||
|
|
@ -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
|
||||
}
|
||||
|
|
|
|||
|
|
@ -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<Integer> 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;
|
||||
|
||||
}
|
||||
|
||||
|
||||
}
|
||||
|
|
|
|||
|
|
@ -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);
|
||||
}
|
||||
}
|
||||
|
|
@ -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<Node> 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);
|
||||
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
|
||||
}
|
||||
Loading…
Reference in New Issue