分治算法 求数组逆序数

题目：在数组中的两个数字如果前面一个数字大于后面一个数字 则这两个数字组成一个逆序对 输入一个数组 求这个数组中逆序对的个数

先把数组分割成子数组 先统计出子数组内部的逆序对的数目 然后再统计出两个相邻子数组之间的逆序对的数目 统计逆序对的过程中 还需要对数组进行排序 这类似与归并排序 算法的时间复杂度为O（nlogn) 空间复杂度为O（n)

归并排序

#include<iostream>using namespace std;void merge(int *A, int *temp, int start, int mid, int end){int a = start;int b = mid+1;int c = end;int d = start;while (a <= mid && b <=end ){if (A[a] < A[b])temp[d++] = A[a++];elsetemp[d++] = A[b++];}while (a != mid + 1)temp[d++] = A[a++];while (b != end + 1)temp[d++] = A[b++];for (int i = start;i <= end;i++)A[i] = temp[i];}void sortg(int *A,int *temp ,int a,int b){if (a < b){int mid = a + (b - a) / 2;sortg(A, temp, a, mid);sortg(A, temp, mid + 1, b);merge(A, temp, a, mid, b);}}int main(){int A[] = { 7,4,10,6,9,9,2,3 };int n = 8;int *t = new int[n];memset(t, 0, sizeof(int)*n);sortg(A, t, 0, n - 1);for (int i = 0;i < n;i++)cout << A[i] << ' ';delete[]t;return 0;}

求逆序对数~~~

#include<iostream>using namespace std;int merge(int *A, int *temp, int start, int mid, int end){int a = end;int b = mid;int c = end;int t = 0;while (b >= start && a >= mid + 1){if (A[b] > A[a]){temp[c--] = A[b--];t += a - mid;}elsetemp[c--] = A[a--];}while (a >= mid + 1)temp[c--] = A[a--];while (b >= start)temp[c--] = A[b--];for (int i = start;i <= end;i++)A[i] = temp[i];return t;}int  Nimum(int *A, int *temp, int a, int b){if (a==b)return 0;else{int mid = a + (b - a) / 2;int left=Nimum(A, temp, a, mid);int right=Nimum(A, temp, mid + 1, b);int num=merge(A, temp, a, mid, b);return left + right + num;}}int main(){int A[] = { 7,4,5,3,8,2};int n =6;int *t = new int[n];memset(t, 0, sizeof(int)*n);int num=Nimum(A, t, 0, n - 1);for (int i = 0;i < n;i++)cout << A[i] << ' ';cout << endl<< num << endl;delete[]t;return 0;}