AcWing算法基础课---第一讲基础算法---01排序

快速排序

步骤

1. 确定分界点：q[l], q[(l+r)/2], q[r], 随机
2. 调整区间
3. 递归处理

void quick_sort(int q[], int l, int r)
{
    if (l >= r) return; //递归结束条件
    
    int i = l - 1, j = r + 1, x = q[l + r >> 1]; //定义i, j指针, 确定分界点x（一般取中间值）
    while (i < j)
    {
        do i ++; while (q[i] < x); 
        do j --; while (q[j] > x);
        if (i < j) swap(q[i], q[j]); //交换两值
    }
    quick_sort(q, l, j), quick_sort(q, j + 1, r);
}

归并排序

步骤

1. 确定分界点 mid
2. 递归排序left, right
3. 归并---合二为一

void merge_sort(int q[], int l, int r)
{
    if (l >= r) return;
    int mid = l + r >> 1; //确定分界点
    merge_sort(q, l, mid);
    merge_sort(q, mid + 1, r);
    
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r)
    {
        if (q[i] < q[j]) temp[k ++] = q[i ++];
        else temp[k ++] = q[j ++];
    }
    while (i <= mid) temp[k ++] = q[i ++];
    while (j <= r) temp[k ++] = q[j ++];
    
    for (int i = l, j = 0; i <= r; i ++, j ++) q[i] = temp[j]; 
}

AcWing 786 第k个数

#include <iostream>

using namespace std;

const int N = 100010; 
int n, k;
int q[N];

void quick_sort(int q[], int l, int r)
{
    if (l >= r) return;
    
    int i = l - 1, j = r + 1, x = q[l + r >> 1];
    while (i < j)
    {
        do i ++; while (q[i] < x);
        do j --; while (q[j] > x);
        if (i < j) swap(q[i], q[j]);
    }
    quick_sort(q, l, j), quick_sort(q, j + 1, r);
}

int main()
{
    cin >> n >> k;
   
    for (int i = 0; i < n; i ++) scanf("%d", &q[i]);
    
    quick_sort(q, 0, n - 1);
    
    cout << q[k - 1];
    
    return 0;
}

AcWing 788. 逆序对的数量

思路
逆序对数量为归并两边的数量和(merge_sort(l, mid) + merge_sort(mid + 1, r))加上当左边值严格大于右边值时的数量(mid - i + 1)

#include <iostream>

using namespace std;

typedef long long LL;

const int N = 100010;

int n;
int q[N], temp[N];


LL merge_sort(int l, int r)
{
    if (l >= r) return 0;
    
    int mid = l + r >> 1;
    LL res = merge_sort(l, mid) + merge_sort(mid + 1, r);
    
    int i = l, j = mid + 1, k = 0;
    while (i <= mid && j <= r)
    {
        if (q[i] <= q[j]) temp[k ++] = q[i ++];
        else
        {
            temp[k ++] = q [j ++];
            res += mid - i + 1;
        }
    }
    //扫尾
    while (i <= mid) temp[k ++] = q[i ++];
    while (j <= r) temp[k ++] = q[j ++];
    //还原原数组
    for (int i = l, j = 0; i <= r; i ++, j ++) q[i] = temp[j];
    
    return res;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> q[i];
    
    LL res = merge_sort(0, n - 1);
    
    cout << res << endl;
    return 0;
}