算法导论第三版2.3答案

2.3-1

划分：
{3}， {41}， {52}， {26}， {38}， {57}， {9}， {49}

合并：
{3， 41}， {26， 52}， {38， 57}， {9， 49}
{3， 26， 41， 52}， {9， 38， 49， 57}
{3， 9， 26， 38， 41， 49， 52，57}

2.3-2

void merge(int a[], int l, int mid, int r)
{
    
      
	int n1 = mid - l + 1, n2 = r - mid;
	for(int i = 1; i <= n1; ++i) L[i] = a[l + i - 1];
	for(int i = 1; i <= n2; ++i) R[i] = a[mid + i];
	int i = 1, j = 1, k = l;
	while(i <= n1 && j <= n2)
	{
    
      
		if(L[i] <= R[j]) a[k ++] = L[i++];
		else a[k ++] = R[j++];
	}
	
	while(i <= n1) a[k ++] = L[i ++];
	while(j <= n2) a[k ++] = R[j ++];
}

2.3-3

$$\begin{gathered} 设n=2^k. \\ 若n=2,k=1,T(n)=2=2lg2,成立. \\ 若n=2^k,k>1,T(n)=nlgn=k\ast 2^k \\ 若n=2^{k+1},k>1 \\ T(n)=(k+1)\ast 2^{k+1} \\ =2T(n/2)+n \\ =2T(2^k)+2^{k+1} \\ =2k\ast 2^k+2^{k+1} \\ =k\ast 2^{k+1}+2^{k+1} \\ =(k+1)\ast 2^{k+1}=nlgn. \end{gathered}$$

2.3-4

$$T(n)=\{\begin{array}{ll}1& n=1\\ T(n-1)+O(n)& n>1\end{array}$$

2.3-5

/*
* @Author: gorsonpy
* @Date:   2022-12-17 15:46:15
* @Last Modified by:   gorsonpy
* @Last Modified time: 2022-12-17 15:46:15
*/
#define NIL -1
int binary_search(int a[], int n, int x)
{
    
      
	int l = 1, r = n;
	while(l < r)
	{
    
      
		int mid = l + r >> 1;
		cout << l << " " << r << " " << mid << endl;
		if(a[mid] < x) l = mid + 1;
		else r = mid;
	}

	if(a[l] != x) return NIL;
	return l;
}

每次都把搜索范围减去一半，故为$O(lgn)$

2.3-6

这题有点不太明白，感觉书上讲的有点问题，二分优化的插入排序虽然确定插入位置只要$O(lgn)$，但是移动还是需要$O(n)$啊。。

void insert_sort(int a[], int n)
{
    
      
	for(int i = 2; i <= n; ++i)
	{
    
      
		int x = a[i];
		a[i] = INF, a[0] = -2e9;
		int l = 0, r = i;
		while(l < r)
		{
    
      
			int mid = l + r >> 1;
			if(a[mid] < x) l = mid + 1;
			else r = mid;
		}
		int j = i - 1;
		while(j && j >= l) a[j + 1] = a[j], --j;
		a[l] = x;
	}
}

2.3-7

这个就是枚举第一个加数是哪个，然后根据大小关系选择在左右的其中一个部分进行二分查找$x-a_i$。二分的代码不好写，瞎写了一个也不知道对不对，反正思路是这么个思路~

/*
* @Author: gorsonpy
* @Date:   2022-12-17 16:19:10
* @Last Modified by:   gorsonpy
* @Last Modified time: 2022-12-17 16:33:57
*/
/*
* @Author: gorsonpy
* @Date:   2022-12-17 16:19:10
* @Last Modified by:   gorsonpy
* @Last Modified time: 2022-12-17 16:37:00
*/
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10;
int a[N];

int main()
{
    
      
	int n, x;
	cin >> n;
	for(int i = 0; i < n; ++i) cin >> a[i];
	cin >> x;
	bool success = false;
	for(int i = 0; i < n; ++i)
	{
    
      
		int key = x - a[i], l, r;
		if(a[i] < key) 
		{
    
      
			l = i + 1, r = n;
			while(l < r)
			{
    
      
				int mid = l + r + 1 >> 1;
				if(a[mid] > key) r = mid - 1;
				else l = mid;
			}
			if(a[l] == key) success = true;
		}
		else if(a[i] > key)
		{
    
      
			l = 0, r = i - 1;
			while(l < r)
			{
    
      
				int mid = l + r >> 1;
				if(a[mid] < key) l = mid + 1;
				else r = mid;
			}
			if(a[r] == key) success = true;
		}
		else
		{
    
      
			l = i + 1, r = n;
			while(l < r)
			{
    
      
				int mid = l + r + 1 >> 1;
				if(a[mid] > key) r = mid - 1;
				else l = mid;
			}
			if(a[l] == key) success = true;
			l = 0, r = i - 1;
			while(l < r)
			{
    
      
				int mid = l + r >> 1;
				if(a[mid] < key) l = mid + 1;
				else r = mid;
			}
			if(a[r] == key) success = true;
		}
	}

	cout << (success ? "YES" : "NO") << endl;
	return 0;
}

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