【挚文集团2025届秋招】Java开发工程师（陌陌业务）-服务端综合-笔试​

【挚文集团2025届秋招】Java开发工程师（陌陌业务）-服务端综合-笔试 ​【挚文集团2025届秋招】Java开发工程师（陌陌业务）-服务端综合-笔试

时间：2024-09-06 19:00

选择 ​编程 ​1 ​代码：

cpp// Description: 【挚文集团2025届秋招】Java开发工程师（陌陌业务）-服务端综合-笔试-编程第一题

// Accept: 100%

// Date: 2024/09/06

#include

#include

using namespace std;

class Node {

public:

char value;

vector children = vector();

explicit Node(char _value) : value(_value) {}

~Node() {

for (Node *node: children) {

delete node;

}

}

void addChild(Node *child) {

if (child && find(children.begin(), children.end(), child) == children.end()) {

children.emplace_back(child);

}

}

[[nodiscard]] pair> longestPath() const {

if (children.empty()) {

return {1, {const_cast(this)}}; // 如果没有子节点，路径长度为 0，路径只有当前节点

}

int maxPathLength = 0;

vector longestPathNodes;

for (const auto &child: children) {

auto [childPathLength, childPathNodes] = child->longestPath();

if (childPathLength > maxPathLength) {

maxPathLength = childPathLength;

longestPathNodes = move(childPathNodes);

}

}

longestPathNodes.insert(longestPathNodes.begin(), const_cast(this)); // 将当前节点添加到路径前面

return {maxPathLength + 1, longestPathNodes};

}

};

class Solution {

private:

vector nodes = vector(26);

vector roots = vector(26);

public:

string LongestBehaviorPath(vector &paths) {

for (const string &path: paths) {

for (int i = 0; i < path.size(); i += 3) {

Node *node = getNodes(path[i]);

if (i == 0) {

if (roots[path[i] - 'A'] == 0) {

roots[path[i] - 'A'] = 1;

}

continue;

}

Node *parent = getNodes(path[i - 3]);

parent->addChild(node);

roots[path[i] - 'A'] = -1;

}

}

string longestPath;

for (int i = 0; i < roots.size(); ++i) {

if (roots[i] != 1) {

continue;

}

auto res = nodes[i]->longestPath().second;

string longest_path;

for (int j = 0; j < res.size() - 1; ++j) {

longest_path += string(1, res[j]->value) + "->";

}

longest_path += string(1, res[res.size() - 1]->value);

if (longest_path.size() > longestPath.size()) {

longestPath = longest_path;

}

}

return longestPath;

}

Node *getNodes(char v) {

if (nodes[v - 'A'] == nullptr) {

nodes[v - 'A'] = new Node(v);

}

return nodes[v - 'A'];

}

};2 ​

代码：

cpp// Description: 【挚文集团2025届秋招】Java开发工程师（陌陌业务）-服务端综合-笔试-编程第二题

// Accept: 100%

// Date: 2024/09/06

#include

#include

using namespace std;

class Solution {

public:

int mergeSort(vector& record, vector& tmp, int l, int r) {

if (l >= r) {

return 0;

}

int mid = (l + r) / 2;

int inv_count = mergeSort(record, tmp, l, mid) + mergeSort(record, tmp, mid + 1, r);

int i = l, j = mid + 1, pos = l;

while (i <= mid && j <= r) {

if (record[i] <= record[j]) {

tmp[pos] = record[i];

++i;

inv_count += (j - (mid + 1));

}

else {

tmp[pos] = record[j];

++j;

}

++pos;

}

for (int k = i; k <= mid; ++k) {

tmp[pos++] = record[k];

inv_count += (j - (mid + 1));

}

for (int k = j; k <= r; ++k) {

tmp[pos++] = record[k];

}

copy(tmp.begin() + l, tmp.begin() + r + 1, record.begin() + l);

return inv_count;

}

int InversionCount(vector &timeSequence) {

int n = timeSequence.size();

vector tmp(n);

return mergeSort(timeSequence, tmp, 0, n - 1);

}

};