最小的k个数

题目描述

　　输入n个整数，找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字，则最小的4个数字是1,2,3,4,

　　解法1：基于Partition函数的时间复杂度为O(n)的解法，只有当我们可以修改原始数组时可用

class Solution {
public:
    int Partition(vector<int> &input, int begin, int end)
    {
        int tmp=input[begin];
        while(begin<end)
        {
            while(begin<end && input[end]>=tmp)--end;
            input[begin]=input[end];
            while(begin<end && input[begin]<=tmp)++begin;
            input[end]=input[begin];
        }
        input[begin]=tmp;
        return begin;
    }
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        int length=input.size();
        vector<int> res;
        if(length==0 || k>length || k<=0)return res;
        int begin=0;
        int end=length-1;
        while(true)
        {
            int mid=Partition(input, begin, end);
            if(mid==k)break;
              if(mid>k)end=mid-1;
            else begin=mid+1;
        }
        for(int idx=0; idx<k; ++idx)
        {
            res.push_back(input[idx]);
        }
        return res;
    }
};

　　解法2：时间复杂度为O(nlogk)的算法。遍历数组，用multiset（底层为红黑树）来存放最小的k个数

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        int length=input.size();
        vector<int> res;
        if(length==0 || k<=0 || k>length)return res;
        std::multiset<int, std::greater<int>> minSet;
        for(int idx=0; idx<length; ++idx)
        {
            if(minSet.size()<k)minSet.insert(input[idx]);
            else{
                std::multiset<int>::iterator it=minSet.begin();
                if(*it>input[idx])
                {
                    minSet.erase(it);
                    minSet.insert(input[idx]);
                }
            }
        }
        for(auto it=minSet.begin(); it!=minSet.end(); ++it)
        {
            res.push_back(*it);
        }
        return res;
    }
};

 　　解法3：时间复杂度为O(nlogk)的算法。遍历数组，用大顶堆来存放最小的k个数

void heapAdjust(vector<int> &heap)
{
    int parent=0;
    int child=2*parent+1;
    while(child<heap.size())
    {
        if(child+1<heap.size() && heap[child+1]>heap[child])++child;
        if(heap[parent]<heap[child])
        {
            int tmp=heap[parent];
            heap[parent]=heap[child];
            heap[child]=tmp;
        }
        parent=child;
        child=2*parent+1;
    }
}
vector<int> GetLeastNumbers_Solution(vector<int> input, int k)
{
    if(input.size()<=k)return input;;
    vector<int> heap;
    if(k<=0)return heap;
    for(int i=0; i<k; ++i)heap.push_back(INT_MAX);
    for(int i=0; i<input.size(); ++i)
    {
        if(input[i]<heap[0])
        {
            heap[0]=input[i];
            heapAdjust(heap);//调整大顶堆
        }
    }
    return heap;
}