PAT甲刷题——U4.9

注意：允许转载，但转载请注明作者和出处

最重要：复习归并排序和快速排序

归并排序

#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int a[N], tmp[N];
int n;

void merge(int a[], int l, int r)
{
    if (l >= r) return;
    int mid = l + r >> 1;
    merge(a, l, mid);
    merge(a, mid + 1, r);
    int i = l, j = mid + 1;
    int k = 0;
    while (i <= mid && j <= r)
    {
        if (a[i] < a[j]) tmp[k ++] = a[i ++];
        else tmp[k ++] = a[j ++];
    }
    while (i <= mid) tmp[k ++] = a[i ++];
    while (j <= r) tmp[k ++] = a[j ++];
    for (int i = l, j = 0; i <= r; i ++, j ++)
        a[i] = tmp[j];
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> a[i];
    merge(a, 0, n - 1);
    for (int i = 0; i < n; i ++) cout << a[i] << ' ';
    puts("");
    return 0;
}

快速排序

#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int a[N];
int n;
void quick_sort(int a[], int l, int r)
{
    if (l >= r) return;
    int x = a[l + r >> 1];
    int i = l - 1, j = r + 1;
    while (i < j)
    {
        do i ++; while(a[i] < x);
        do j --; while(a[j] > x);
        if (i < j) swap(a[i], a[j]);
    }
    quick_sort(a, l, j);
    quick_sort(a, j + 1, r);
}
int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> a[i];
    quick_sort(a, 0, n - 1);
    for (int i = 0; i < n; i ++) cout << a[i] << ' ';
    puts("");
    return 0;
}

AcWing 1579. 插入还是归并

题目

答案

2024年冗杂版

// 个人感觉这个答案好理解一些。。。。
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int a[N], b[N];
int n;
bool check(int a[], int b[])
{
    for (int i = 0; i < n; i ++)
        if (a[i] != b[i]) return false;
    return true;
}
int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> a[i];
    for (int i = 0; i < n; i ++) cin >> b[i];
    
    int p = 1;
    while (p < n && b[p] >= b[p - 1]) p ++;
    int k = p;
    while (k < n && a[k] == b[k]) k ++;
    // cout << k << endl;
    if (k == n)
    {
        puts("Insertion Sort");
        while (p && b[p - 1] > b[p])
        {
            swap(b[p - 1], b[p]);
            p --;
        }
        for (int i = 0; i < n; i ++) cout << b[i] << ' ';
        puts("");
    }
    else
    {
        puts("Merge Sort");
        int step = 1;
        while (true)
        {
            int len = 1 << step;
            for (int i = 0; i < n; i += len)
            {
                int j = min(i + len, n);
                sort(a + i, a + j);
            }
            // for (int i = 0; i < n; i ++) cout << a[i] << ' ';
            // puts("");
            if (check(a, b)) break;
            step ++;
        }
        step ++;
        int len = 1 << step;
        for (int i = 0; i < n; i += len)
        {
            int j = min(i + len, n);
            sort(a + i, a + j);
        }
        for (int i = 0; i < n; i ++) cout << a[i] << ' ';
        puts("");
    }
    
    return 0;
}

2022年 简易版

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int a[N], b[N];
int n;

bool check()
{
    for (int i = 0; i < n; i ++)
        if (a[i] != b[i]) return false;
    
    return true;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> a[i];
    for (int i = 0; i < n; i ++) cin >> b[i];
    
    int k = 0;
    while (b[k + 1] >= b[k]) k ++;
    
    bool match = true;
    for (int i = k + 1; i < n; i ++)
    {
        if (a[i] != b[i])
        {
            match = false;
            break;
        }
    }
    
    if (match)  // 如果是插入排序
    {
        puts("Insertion Sort");
        sort(b, b + k + 2);
        for (int i = 0; i < n; i ++) cout << b[i] << ' ';
        puts("");
    }
    else    // 如果是归并排序，按步长为1，2，4，8，16这样判断，归并排序特点是是步长内有序
    {
        puts("Merge Sort");
        
        int step = 1;
        while (true)
        {
            bool match = check();   // a和b匹配了吗？不管匹不匹配，都得继续往下步长内排序
            
            int len = 1 << step;   // 左移k位，相当于 1 * 2^k
            for (int i = 0; i < n; i += len)    // 步长内排序
            {
                int j = min(i + len, n);
                sort (a + i, a + j);
            }
            
            if (match) break;
            step ++;
        }
        
        for (int i = 0; i < n; i ++) cout << a[i] << ' ';
    }
    
    return 0;
}