LeetCode DAY11(239. Sliding Window Maximum&347. Top K Frequent Elements)

Preface

This is a new day to continue my stack and queue journey.
Learn something new and keep reviewing what I learnt before.

1. Sliding Window Maximum

LeetCode Link: 239. Sliding Window Maximum
You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.

Return the max sliding window.

Analysis and Solution

Sliding Window

LeetCode C++ as followings Sliding Window

class Solution {
private:
    class MyQueue { //monotonous queue (from less to large)
    public:
        deque<int> que; // double enden queue realize monotonous queue 
        void pop(int value) {
            if (!que.empty() && value == que.front()) {//compare value need to be poped with value in the exit of the queue. pop former if former is equal to latter; queue should not be empty.
                que.pop_front();
            }
        }
        void push(int value) {
            while (!que.empty() && value > que.back()) {//value to be pushed is larger than value in the entrance of the queue. pop back all the values in the back of the queue until value to be pushed is less than or equal to value in the entrance of the queue.
                que.pop_back();
            }
            que.push_back(value);//repush the suitable value 

        }
        int front() {//the max avlue is the value in the front of the queue
            return que.front();
        }
    };
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        MyQueue que;
        vector<int> result;
        for (int i = 0; i < k; i++) { // traverse first K elements, import them into queue
            que.push(nums[i]);
        }
        result.push_back(que.front()); // result records the max value of the first K elements
        for (int i = k; i < nums.size(); i++) {
            que.pop(nums[i - k]); // sliding window pops the front element
            que.push(nums[i]); // slidingwindow pushes the value in the back
            result.push_back(que.front()); // record the max value
        }
        return result;
    }
};

2.Top K Frequent Elements

LeetCode Link: 347. Top K Frequent Elements
Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Analysis and Solution

Priority Queue

LeetCode C++ as followings Priority Queue

class Solution {
public:
    // heap
    class mycomparison {//self define function
    public:
        bool operator()(const pair<int, int>& lhs, const pair<int, int>& rhs) {
            return lhs.second > rhs.second;
        }
    };
    vector<int> topKFrequent(vector<int>& nums, int k) {
        // count frequency of the elements
        unordered_map<int, int> map; // map
        for (int i = 0; i < nums.size(); i++) {//traverse nums
            map[nums[i]]++;
        }

        // sorting for frequency
        // define a heap , size is k
        priority_queue<pair<int, int>, vector<pair<int, int>>, mycomparison> pri_que;

        //  use the heap to scan the value of the frequency
        for (unordered_map<int, int>::iterator it = map.begin(); it != map.end(); it++) {
            pri_que.push(*it);
            if (pri_que.size() > k) { // pop if the size of heap is larger than K; gurante the size is K
                pri_que.pop();
            }
        }

        // find the first K frequent elements; print value in reverse order
        vector<int> result(k);
        for (int i = k - 1; i >= 0; i--) {//
            result[i] = pri_que.top().first;
            pri_que.pop();
        }
        return result;

    }
};