归并排序 逆序对的实现

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归并排序 逆序对的实现

#include<iostream>
using namespace std;

long long mergeSort(int a[], int n);
long long _mergeSort(int a[], int l, int r);
long long merge(int a[], int l, int mid, int r);

void test_mergeSort();
void test_mergeSort2();
void test_mergeSort3();

int main(){
    test_mergeSort();
    test_mergeSort2();
    test_mergeSort3();

    return 0;
}

//////////////////////////////////////////////////////////////////////////
// 归并排序
// 计算逆序数对的结果以long long返回
// 对于一个大小为N的数组, 其最大的逆序数对个数为 N*(N-1)/2, 非常容易产生整型溢出
// 逆序对个数最多不超过：(n-1)+(n-2)+(n-3)+...+2+1=n*(n-1)/2;

long long mergeSort(int a[], int n){
    return _mergeSort(a, 0, n - 1);
}
long long _mergeSort(int a[], int l, int r){
    if (l >= r) return 0;
    int mid = l + (r - l) / 2;
    long long result_l = _mergeSort(a, l, mid);//a[l...mid]范围内的逆序对
    long long result_r = _mergeSort(a, mid + 1, r);//a[mid+1...r]范围内的逆序对
    return result_l + result_r + merge(a, l, mid, r);
}
long long merge(int a[], int l, int mid, int r){
    int *aux = new int[r - l + 1];//辅助数组
    for (int i = l; i <= r; ++i)
        aux[i - l] = a[i];

    long long result = 0;//逆序对个数

    int i = l, j = mid + 1;//左右块的起始索引分别为i,j
    for (int k = l; k <= r; ++k){
        if (i > mid){ a[k] = aux[j - l]; j++; }//左块索引超出索引范围
        else if (j > r){ a[k] = aux[i - l]; i++; }//右块索引超出索引范围
        else if (aux[i - l] <= aux[j - l]){ a[k] = aux[i - l]; i++; }//左半部分 <= 右半部分
        else { a[k] = aux[j - l]; j++;//左半部分 > 右半部分
            result += (long long)(mid - i + 1);//右半部分元素小，故j所指的右半部分的当前元素a[j] < 左半部分当前元素a[i]以及左半部分i之后的所有剩余元素，那么当前逆序对个数就是 mid-i+1; [l...i...mid; mid+1...j...r]
        }
    }
    delete[] aux;//删掉new的辅助数组aux

    return result;
}

//////////////////////////////////////////////////////////////////////////三个测试用例
void test_mergeSort(){
    int a[] = { 4, 36, 12, 3, 16, 5, 7, 0, 12, 4, 16 };
    int n = sizeof(a) / sizeof(a[0]);

    for (int i = 0; i < n; ++i)
        cout << a[i] << "\t";
    cout << endl;

    cout << "逆序对个数为：" << mergeSort(a, n) << endl;

    for (int i = 0; i < n; ++i)
        cout << a[i] << "\t";
    cout << endl;
}

void test_mergeSort2(){
    int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int n = sizeof(a) / sizeof(a[0]);

    for (int i = 0; i < n; ++i)
        cout << a[i] << "\t";
    cout << endl;

    cout << "逆序对个数为：" << mergeSort(a, n) << endl;

    for (int i = 0; i < n; ++i)
        cout << a[i] << "\t";
    cout << endl;
}

void test_mergeSort3(){
    int a[] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };
    int n = sizeof(a) / sizeof(a[0]);

    for (int i = 0; i < n; ++i)
        cout << a[i] << "\t";
    cout << endl;

    cout << "逆序对个数为：" << mergeSort(a, n) << endl;

    for (int i = 0; i < n; ++i)
        cout << a[i] << "\t";
    cout << endl;
}