go快排计算最小k个数和第k大的数

计算最小的k个数，除了用堆，还可以用快排

每次返回left right相遇的位置，index>k 则向左搜，index<k 向右搜，直到index==k

剑指40 最小k个数（没要求顺序）

• golang快排

func getLeastNumbers(arr []int, k int) []int {
    if k==0{
        return []int{}
    }
    index:=Fast(arr,0,len(arr)-1,k)
    for index!=k-1{
        if index>k-1{
            index = Fast(arr,0,index-1,k)
        }else{
            index = Fast(arr,index+1,len(arr)-1,k)
        }
    }
    res:=make([]int,index+1)
    copy(res,arr[:index+1])
    return res
}

func Fast(nums []int,low,high,k int)int {
    start:=low
    end:=high
    tmp:=nums[low]

    for start<end{
        for start<end&&nums[end]>=tmp{
            end--
        }
        nums[start] = nums[end]

        for start<end&&nums[start]<=tmp{
            start++
        }
        nums[end] = nums[start]
    }
    nums[start] = tmp
    return start
}
 

• c++版本

 vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        if(input.size()<k||k==0) return {};
       
        int index = fast(input,0,input.size()-1);
        while(index!=k-1)
        {
            if(index>k-1)
                index = fast(input,0,index-1);
            else
                index = fast(input,index+1,input.size()-1);
        }
        return vector<int>(input.begin(),input.begin()+k);
    }
    int fast(vector<int>&arr,int low,int high)
    {
        if(low>=high)return low;
        int start = low;
        int end = high;
        int tmp = arr[low];
        
        while(start<end)
        {
            while(start<end&&arr[end]>=tmp) end--;
            arr[start] = arr[end];
            while(start<end&&arr[start]<=tmp) start++;
            arr[end] = arr[start];
        }
        arr[start] = tmp;
        return start;
    }

leetcode 215 第k个最大元素

func findKthLargest(nums []int, k int) int {
    return Fast(nums,0,len(nums)-1,k)
}

func Fast(nums []int,low,high,k int)int  {

    start:=low
    end:=high
    tmp:=nums[low]
    for start<end{
        for start<end&&nums[end]<=tmp{
            end--
        }
        nums[start] = nums[end]
        for start<end&&nums[start]>=tmp{
            start++
        }
        nums[end] = nums[start]
    }
    nums[start] = tmp
    if start==k-1{
        return nums[k-1]
    }else if start>k-1{
        return Fast(nums,low,start-1,k)
    }else{
        return Fast(nums,start+1,high,k)
    }
}