找出最小的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,。

Solution

使用快排解决，基于数组的第k个数字来调整，使得比第k个数字小的所有数字都位于数组的左边，比第k个数字大的所有数字都位于数组的右边。调整之后，位于数组左边的k个数字就是最小的k个数字（这k个数字不一定是排序的）。O(N)

# -*- coding:utf-8 -*-
class Solution:
    def GetLeastNumbers_Solution(self, tinput, k):
        # write code here
        def quick_sort(lst):
            if not lst:
                return []
            pivot = lst[0]
            left = quick_sort([x for x in lst[1: ] if x < pivot])
            right = quick_sort([x for x in lst[1: ] if x >= pivot])
            return left + [pivot] + right
 
        if tinput == [] or k > len(tinput):
            return []
        tinput = quick_sort(tinput)
        return tinput[: k]

class Solution {
public:
    void swap(int &fir,int &sec)
    {
        int temp = fir;
        fir = sec;
        sec = temp;
    }
     
  int getPartition(vector<int>  a, int left, int right)
{
    int x = a[right];
    int i = left-1, j = right;
    for (;;)
    {
        while(a[++i] < x) { }
        while(a[--j] > x) { if(j==left) break;}
        if(i < j) 
            swap(a[i], a[j]);
        else break;
    }
    swap(a[i],a[right]);
    return i;
}

    vector<int> GetLeastNumbers_Solution(vector<int> input, int k)
    {
        vector<int> result;       
        if(input.empty() || k>input.size() || k<=0) return result;
         
        int start = 0;
        int end = input.size()-1;
        int index = getPartition(input,start,end);
         
        while(index != (k-1))
        {
            if(index > (k-1))
            {
                end = index - 1;
                index = getPartition(input,start,end);
            }
            else
            {
                start = index + 1;
                index = getPartition(input,start,end);
            }
        }
         
        for(int i=0;i<k;++i)
        {
            result.push_back(input[i]);
        }
         
        return result;
    }
};