TopK问题

题目描述

下面介绍三种方法

一、暴力法

class Solution {
public:
    vector<int> getLeastNumbers(vector<int>& arr, int k) {
        sort(arr.begin(),arr.end(), less<int>());
        vector<int> res(arr.begin(),arr.begin()+k);
        return res;
    }
};

二、快排划分思想

class Solution {
public:
    int partition(vector<int> &arr, int low, int high){
        int value = arr[low];
        while(low < high){
            //注意这里必须加等号，否则arr[low]==arr[high] 且 arr[low]==value会有死循环
            while(arr[high]>=value && low<high) high--;
            arr[low] = arr[high];
            while(arr[low]<=value && low<high) low++;
            arr[high] = arr[low];
        }
        arr[low] = value;
        return low;
    }

    vector<int> getLeastNumbers(vector<int>& arr, int k) {
        if(k == 0) return vector<int>();
        else if(k >= arr.size()) return arr;
        else{
            int low = 0;
            int high = arr.size() - 1;
            int index = partition(arr, low, high);
            while(index != k-1){
                if(index < k-1) low = index + 1;
                else high = index - 1;
                index = partition(arr, low, high);
            }
            return vector<int> (arr.begin(), arr.begin()+k);
        }
    }
};

三、优先级队列

优先级队列本质是通过堆实现的

• less 表示大顶堆
• greater 表示小顶堆

class Solution {
public:
    vector<int> getLeastNumbers(vector<int>& arr, int k) {
        vector<int> res(k);
        priority_queue<int,vector<int>,less<int>> q;
        for(int i=0;i<arr.size();i++){
            q.push(arr[i]);
            if(q.size()>k) q.pop();
        }
        for(int i=k-1;i>=0;i--){
            res[i]=q.top();
            q.pop();
        }
        return res;
    }
};

四、堆排序

​​堆排序详解​​