什么是逆序对?数组中任意两个元素组合是降序排列的,以下给出了暴力求解法和归并求解法。
#include<iostream> using namespace std; #define N 12 //逆序对:其实就是求每个数右边有多少个数比其小 //暴力求解法 int VolentReversePair(int *a,int length) { int cnt=0; for(int i=0;i<length;i++) { for(int j=i;j<length;j++) { if(a[j] < a[i]) cnt ++; } } return cnt; } int Merge(int *a,int left,int middle,int right) { int sum=0; int i=left; int j=middle+1; int k=0; int *buf = new int[right-left+1]; while(i <= middle && j<=right) { if(a[i] <= a[j]) { buf[k++] = a[i++]; } else { sum += middle -i+1; buf[k++] = a[j++]; } } while(i<=middle) { buf[k++] = a[i++]; } while(j<=right) { buf[k++] = a[j++]; } for(int m=0,n=left;m<right-left+1;m++,n++) { a[n] = buf[m]; } delete[] buf; return sum; } int mergeSort(int *a, int left, int right) { if(left == right) return 0; int middle = left + ((right - left)>>1); return mergeSort(a,left,middle) + mergeSort(a,middle+1,right)+Merge(a,left,middle,right); } int main() { int input[N]={5,4,3,2,1,7,9,8,0,11,15,21}; cout<<VolentReversePair(input,N)<<endl; cout<<mergeSort(input,0,N-1)<<endl; return 0; }