剑指offer: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,。

方法一

O（n）

改变输入，适合小数据

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {

        int len = input.size();
        vector<int> res;

        //注意k<=0的异常输入
        if (k <= 0 || len < k){
            return res;
        }

        int left = 0, right = len - 1;
        int index = partition(input, left, right);

        while (index != k - 1){
            if (index < k - 1){
                left = index + 1;
                index = partition(input, left, right);
            }
            else{
                right = index - 1;
                index = partition(input, left, right);
            }
        }

        for (int i = 0; i <= k - 1; i++){
            res.push_back(input[i]);
        }

        return res;
    }
private:
    int partition(vector<int> &input, int left, int right){
        int i = left, j = right;
        int pivot = input[i];

        while (i < j){
            while (i<j && input[j]>pivot) j--;
            if (i < j){
                input[i++] = input[j];
            }

            while (i < j&&input[i] <= pivot) i++;
            if (i < j){
                input[j--] = input[i];
            }
        }

        input[i] = pivot;

        return i;
    }

};

方法二

大顶堆，适合于大数据情形

class Solution {
public:
    vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {

        vector<int> res;

        int len = input.size();

        if (k <= 0 || k > len){
            return res;
        }

        priority_queue<int> heap;

        for (int i = 0; i < len; i++){
            if (heap.size() < k){
                heap.push(input[i]);
            }
            else{
                if (input[i] < heap.top()){
                    heap.pop();
                    heap.push(input[i]);
                }
            }
        }


        while (!heap.empty()){
            res.push_back(heap.top());
            heap.pop();
        }

        return res;
    }
};