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);
}
};
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