diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection.java index 20e0c32..c847826 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection.java @@ -1,5 +1,7 @@ package com.markilue.leecode.hot100.interviewHot.union_find.second; +import org.junit.Test; + import java.util.ArrayList; /** @@ -12,6 +14,12 @@ import java.util.ArrayList; */ public class LC_685_FindRedundantConnection { + @Test + public void test() { + int[][] edges = {{1, 2}, {1, 3}, {2, 3}}; + System.out.println(findRedundantDirectedConnection(edges)); + } + int[] father;//父节点 int n;//父节点个数 @@ -84,7 +92,7 @@ public class LC_685_FindRedundantConnection { ArrayList twoDegree = new ArrayList<>(); //判断入度为2的节点，该节点一定有子节点需要删除;反向遍历,因为后面的删除优先级更高 - for (int i = edges.length-1; i >=0; i--) { + for (int i = edges.length - 1; i >= 0; i--) { if (inDegree[edges[i][1]] == 2) { twoDegree.add(i);//这个节点需要删除 } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection1.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection1.java new file mode 100644 index 0000000..a79e837 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/second/LC_685_FindRedundantConnection1.java @@ -0,0 +1,109 @@ +package com.markilue.leecode.hot100.interviewHot.union_find.second; + +import org.junit.Test; + +import java.util.ArrayList; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.interviewHot.union_find.second + *@Author: markilue + *@CreateTime: 2023-06-12 09:58 + *@Description: TODO + *@Version: 1.0 + */ +public class LC_685_FindRedundantConnection1 { + + @Test + public void test() { + int[][] edges = {{1, 2}, {1, 3}, {2, 3}}; + System.out.println(findRedundantDirectedConnection(edges)); + } + + int[] father; + + private void init(int[] father) { + for (int i = 0; i < father.length; i++) { + father[i] = i; + } + } + + private int find(int u) { + if (father[u] == u) return u; + father[u] = find(father[u]); + return father[u]; + } + + private void union(int u, int v) { + u = find(u); + v = find(v); + if (u == v) return; + father[v] = u; + } + + private boolean same(int u, int v) { + u = find(u); + v = find(v); + return u == v; + } + + private int[] removeOne(int[][] edges) { + init(father); + for (int[] edge : edges) { + if (same(edge[0], edge[1])) { + return edge; + } + union(edge[0], edge[1]); + + } + return null; + } + + private boolean removeIfCan(int[][] edges, int i) { + init(father); + //遇上i就跳过 + for (int i1 = 0; i1 < edges.length; i1++) { + if (i1 == i) continue; + if (same(edges[i1][0], edges[i1][1])) { + return false; + } + + union(edges[i1][0], edges[i1][1]); + + } + return true; + } + + + //虽然是有向图，但是一共就三种情况，两种入度为2的情况可以直接判断出来，最后可以转为无向图的情况 + public int[] findRedundantDirectedConnection(int[][] edges) { + father = new int[1010]; + //判断入度为2的情况 + int[] countDegree = new int[1010]; + + for (int[] edge : edges) { + countDegree[edge[1]]++; + } + + ArrayList twoDegree = new ArrayList<>();//入度为2的节点 + + for (int i = edges.length - 1; i >= 0; i--) { + if (countDegree[edges[i][1]] > 1) twoDegree.add(i); + } + + if (!twoDegree.isEmpty()) { + if (removeIfCan(edges, twoDegree.get(0))) { + return edges[twoDegree.get(0)]; + } else { + return edges[twoDegree.get(1)]; + } + } + + //只用删除一个 + return removeOne(edges); + + + } + + +} diff --git a/Leecode/src/main/java/com/markilue/leecode/interview/huawei/T0412/Question1.java b/Leecode/src/main/java/com/markilue/leecode/interview/huawei/T0412/Question1.java new file mode 100644 index 0000000..f527fef --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/interview/huawei/T0412/Question1.java @@ -0,0 +1,54 @@ +package com.markilue.leecode.interview.huawei.T0412; + +import java.util.Arrays; +import java.util.Scanner; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.interview.huawei.T0412 + *@Author: markilue + *@CreateTime: 2023-06-12 11:38 + *@Description: + * TODO 交易系统的降级策略: + * 有一个核心交易系统接口被N个上游系统调用，每个上游系统的调用量R=[R1,R2.....,RN]. + * 由于核心交易系统集群故障，需要暂时系统降级限制调用，核心交易系统能接受的最大调用量为cnt。 + * 设置降级规则如下： + * 如果sum(R1.R2..RN)小于等于cnt，则全部可以正常调用，返回-1; + * 如果sum(R1.R2....RN)大于cnt，设置一个阈值limit， + * 如果某个上游系统发起的调用量超过limit，就将该上游系统的调用量限制为limit， + * 其余未达到limit的系统可以正常发起调用。 + * 求出这个最大的limit (limit可以为0) + * 此题目对效率有要求，请选择高效的方式。 + *@Version: 1.0 + */ +public class Question1 { + + public static void main(String[] args) { + Scanner sc = new Scanner(System.in); + int[] nums = Arrays.stream(sc.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray(); + int threshold = Integer.parseInt(sc.nextLine()); + + solve(nums, threshold); + } + + //二分寻找最大值 + private static void solve(int[] nums, int threshold) { + int max = Math.min(threshold,(int) 1e5); + int min = threshold / nums.length; + while (min < max) { + int mid = min + ((max - min + 1) >> 1); + if (check(nums, mid, threshold)) min = mid; + else max = mid - 1; + } + System.out.println(min); + } + + private static boolean check(int[] nums, int max, int threshold) { + int result = 0; + for (int i = 0; i < nums.length; i++) { + if (nums[i] < max) result += nums[i]; + else result += max; + } + return result <= threshold; + } +}