topk两种解法

 1.这个通过partition实现topk，时间复杂度是o(logn*logn)，也就是0(n),但需要修改原数组的顺序

下面这个代码本身有一些错误，并且throw excption会在牛客上报错

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        vector<int> result;
        int length = input.size();
        if(input.empty() || length <= 0 || k <= 0 || length < k)
            return result;
        int start = 0;
        int end = length - 1;
        int index = partition(input,length,start,end);
        while(index != k-1){
            if(index > k-1){
                end = index-1;
                index = partition(input,length,start,end);
            }
            else{
                start = index + 1;
                index = partition(input,length,start,end);
            }
        }
        for(int i = 0;i <= index;i++)
            result.push_back(input[i]);
        return result;
    }
    int partition(vector<int> &input,int length,int start,int end){
        if(input.empty() || length <= 0 || start < 0 || end >= length)
            throw new exception("Invalid Parameters");
        int small = -1;
        for(int i = start;i <= end;i++){
            if(input[i] <= input[end]){
                start++;
                if(start != i)
                    swap(&input[start],&input[i]);
            }
        }
        small++;
        swap(&input[small],&input[end]);
        return small;
    } 
};

 正确代码

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {
        vector<int> result;
        int length = input.size();
        if(input.empty() || length <= 0 || k <= 0 || length < k)
            return result;
        int start = 0;
        int end = length - 1;
        int index = partition(input,start,end);
        while(index != k-1){
            if(index > k-1){
                end = index - 1;
                index = partition(input,start,end);
            }
            else{
                start = index + 1;
                index = partition(input,start,end);
            }
        }
        for(int i = 0;i < k;i++)
            result.push_back(input[i]);
        return result;
    }
    int partition(vector<int> &input,int start,int end){
        int small = start - 1;
        for(int i = start;i < end;i++){
            if(input[i] < input[end]){
                small++;
                if(small != i)
                    swap(input,small,i);
            }
        }
        small++;
        swap(input,small,end);
        return small;
    } 
    void swap(vector<int>& input,int a,int b){
        int tmp = input[a];
        input[a] = input[b];
        input[b] = tmp;
    }
};

2.用大根堆的做法的时间复杂度是o(nlogk)