题解 | #数组中的逆序对#
数组中的逆序对
https://www.nowcoder.com/practice/96bd6684e04a44eb80e6a68efc0ec6c5
class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param nums int整型vector * @return int整型 */ const int kmod = 1000000007; int InversePairs(vector<int>& nums) { // write code here vector<int> tmp(nums.size()); int res=0; mergeSort(nums,0,nums.size()-1,res,tmp); return res; } void merge(vector<int> &nums,int left,int mid,int right,int &res,vector<int> &tmp){ //if(left==right) return ; int i=left,j=mid+1,k=0; while(i<=mid&&j<=right){ if(nums[i]<=nums[j]){ tmp[k++]=nums[i++]; }else{ tmp[k++]=nums[j++]; res+=(mid-i+1); res%=kmod; } } while(i<=mid) tmp[k++]=nums[i++]; while(j<=right) tmp[k++]=nums[j++]; //tmp是复用的!!!!!!! for(int i=left,k=0;i<=right;i++,k++){ nums[i]=tmp[k]; } } void mergeSort(vector<int> &nums,int left,int right,int &res,vector<int> &tmp){ if(left>=right) return; int mid=left+(right-left)/2; mergeSort(nums, left, mid, res,tmp); mergeSort(nums,mid+1,right,res,tmp); merge(nums,left,mid,right,res,tmp); } };