Just Sort It（快速排序）

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#快速排序

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本文介绍了一种需要实现nlogn复杂度的排序算法挑战，并推荐使用快速排序或堆排序来应对严格的内存限制。提供了标准快速排序算法的C语言实现示例。

Judge Info

Memory Limit: 1124KB
Case Time Limit: 10000MS
Time Limit: 10000MS
Judger: Number Only Judger

Description

Sorting is one of the most important algorithm in computer sciences. Please learn it well.

本题需要的排序算法复杂度应该为nlogn，冒泡、选择的负责度为n^2，快排、归并、堆排序的复杂度为nlogn

本题的内存限制也是比较严格，所以建议大家使用堆排序，或者快速排序

Input

The first line of input contains $T(1\leq T \leq 10)$ , the number of test cases. There is only line for each test case. The first number $N(1\leq N \leq 250,000)$ of each test case will indicates how many integer numbers are there for sort from smaller to larger. All numbers are in the range of $[-2^{31},2^{31}-1]\,$ .

Output

Output one line for every test case. The result of those integer numbers after sorting.

Sample Input

Sample Output

1 2 3
2 3
3 4 5

标准快排

#include<stdio.h>

void quicksort(int *s, int l, int r)
{
    if (l < r)
    {
        int i = l, j = r, x = s[l];
        while (i < j)
        {
            while(i < j && s[j] >= x) j--;  
            if(i < j) s[i++] = s[j];
			
            while(i < j && s[i] < x) i++;  
            if(i < j) s[j--] = s[i];
        }
        s[i] = x;
        quicksort(s, l, i - 1); 
        quicksort(s, i + 1, r);
    }
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		int a[250000];
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
		quicksort(a,0,n-1);
		for(int i=0;i<n;i++)
		{
			printf("%d",a[i]);
			if(i<n-1)
			putchar(' ');
		}
		putchar('\n');
	}
	return 0;
}