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,。
好吧初看这个题以为还有什么高级的算法,其实第一眼想到的就是排序,在取前面几个数,实现如下:
#pragma once
#include <vector>
using namespace std;
namespace min_key_number
{
class Solution {
public:
vector<int> GetLeastNumbers_Solution(vector<int> input, int k)
{
vector<int> ret;
if (input.size() < k)
return ret;
quick_sort(input, 0, input.size() - 1);
ret.resize(k);
for (int i = 0; i < k; i++)
{
ret[i] = input[i];
}
return ret;
}
void quick_sort(vector<int> &input, int left, int right)
{
if (left >= right)
return;
int mid_num = input[left];
int mid;
int l = left;
int r = right;
while (left < right)
{
while (left < right && input[left] <= mid_num)
{
left++;
}
while (left < right && input[right] >= mid_num)
{
right--;
}
int exp = input[left];
input[left] = input[right];
input[right] = exp;
}
int exp = input[left - 1];
input[left - 1] = input[l];
input[l] = exp;
quick_sort(input, l, left -1);
quick_sort(input, left + 1, r);
}
};
}
test.cpp
#pragma once
#include "min_key_number.h"
namespace min_key_number
{
class min_key_number_test
{
public:
void start_test()
{
Solution s;
std::vector<int> v;
v.push_back(4);
v.push_back(5);
v.push_back(1);
v.push_back(6);
v.push_back(2);
v.push_back(7);
v.push_back(3);
v.push_back(8);
s.GetLeastNumbers_Solution(v, 3);
}
};
}
好吧吐槽一下,史上最垃圾的快速排序写法。
这个题本质也是一个排序问题,可以采取堆调整的方法,创建一个大根堆,每次移除root节点,同时调整推的顺序也能解决,时间复杂度也是nlogn,同时其他各种排序也能解决