美团笔试

算法第一题



example

input: 
4
1 2
1 3
3 4

output:
4

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

int maxRoadLength(unordered_map<int, vector<int>> &adjList, int root) {
    vector<int> successors = adjList[root];
    size_t size = successors.size();
    if (size == 0) return 0;
    int maxLen = 0;
    for (int successor : successors) {
        maxLen = max(maxLen, maxRoadLength(adjList, successor));
    }
    return maxLen + 1;
}

int main() {
    int N;
    cin >> N;
    unordered_map<int, vector<int>> adjList;
    int pioneer, successor;
    for (int i = 0; i < N-1; i++) {
        cin >> pioneer >> successor;
        adjList[pioneer].push_back(successor);
    }
    int ans = maxRoadLength(adjList, 1);
    ans = 2 * (N-1) - ans;
    cout << ans << endl;
    return 0;
}

算法第二题

这个是一个考察队列的题, 很精髓, 下次遇到这种顺序求解的题, 要注意



example

input:
10 2
1 0 0 1 0 1 0 1 0 1

output: 5

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int main() {
    int N, K;
    cin >> N >> K;
    vector<int> nums(N);
    for (int i = 0; i < N; i++) {
        cin >> nums[i];
    }
    // 连续的 1 串, 必定连续
    // 看从 (start, end] 区间连续的 1 串的长度
    queue<int> zero_queue;
    int max_length = 0;
    int cur_length = 0;
    for(int end = 0; end < N; end++) {
        if (nums[end] == 1) {
            cur_length++;
            max_length = max(max_length, cur_length);
            continue;
        }

        if (zero_queue.size() < K){
            zero_queue.push(end);
            cur_length++;
            max_length = max(max_length, cur_length);
        } else {
            int start = zero_queue.front();
            zero_queue.pop();
            zero_queue.push(end);
            cur_length = end-start;
        }
    }
    cout << max_length << endl;
    return 0;
}