传统求一组数据内次最大和次大元素有顺序搜索法(时间复杂度O(n)),排序法(O(n*logn))等。而分治法可以把时间复杂度降低到O(logn)级别,但是相对来说实现起来也复杂一点

code:

#include <algorithm>
#include <iostream>

using namespace std;
const int INF = 0x3f3f3f3f;

void solve(int a[], int left, int right, int &max1, int &max2)
{
	if(left == right)	// 递归出口,区间内只有一个元素 
	{
		max1 = a[left];
		max2 = -INF;
	}
	else if(left + 1 == right)	// 递归出口,区间内有两个元素 
	{
		max1 = max(a[left], a[right]);
		max2 = min(a[right], a[right]);
	}
	else
	{
		int mid = (left + right) / 2;
		int lmax1, lmax2;
		int rmax1, rmax2;
		solve(a, left, mid, lmax1, lmax2); 	// 递归寻找左边 
		solve(a, mid+1, right, rmax1, rmax2);	// 递归寻找右边 
		if(lmax1 > rmax1)
		{
			max1 = lmax1;
			max2 = max(lmax2, rmax1);
		}
		else
		{
			max1 = rmax1;
			max2 = max(lmax1, rmax2);
		}
	}
}

int main()
{
	int n;
	int a[1000];
	cin >> n;
	for(int i = 0; i < n; i++)
		cin >> a[i];
	int max1, max2;
	solve(a, 0, n-1, max1, max2);
	cout << max1 << " " << max2 << endl;
	
	
	
	return 0;
 }