逆序对问题总结

背景

分析

Brute Force

vector< vector<int> > fun1(vector<int> nums){
    vector< vector<int> > ans;
    int n=nums.size();
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            if(nums[j]<nums[i]){
                ans.push_back({nums[i],nums[j]});
            }
        }
    }
    return ans;
}

BIT

int lowbit(int x){
    return x&(-x);
}

int sum(int x,vector<int>& C){
    int ans=0;
    while(x>0){
        ans+=C[x];
        x-=lowbit(x);
    }
    return ans;
}

void add(int x,int val, int Max, vector<int>& C){
    while(x<=Max){
        C[x]+=val;
        x+=lowbit(x);
    }
}

int fun2(vector<int> nums){
    int n=nums.size();
    unordered_map<int,int> mp;
    vector<int> temp(nums);
    sort(temp.begin(),temp.end());
    int tot=1;
    for(auto i:temp){
        mp[i]=tot++;
    }
    int ans=0;
    vector<int> C(tot+1,0);
    for(int i=n-1;i>=0;i--){
        ans+=sum(mp[nums[i]]-1,C);
        add(mp[nums[i]],1,tot,C);
    }
    return ans;
}

Merge sort

void merge(vector<int>& nums,int l, int mid,int r){
    int n1=mid-l+1;
    int n2=r-mid;

    vector<int> L1;
    vector<int> L2;
    L1.assign(nums.begin()+l,nums.begin()+l+n1);
    L2.assign(nums.begin()+n1+l,nums.begin()+ n1+n2+l);
    int i1=0;
    int i2=0;
    while(i1<n1&&i2<n2){
        if(L1[i1]<=L2[i2]){
            nums[i1+i2+l]=L1[i1];
            i1++;
        } else {
            nums[i1+i2+l]=L2[i2];
            i2++;
        }
    }
    while(i1<n1){
        nums[i1+i2+l]=L1[i1];
            i1++;
    }
    while(i2<n2){
        nums[i1+i2+l]=L2[i2];
            i2++;
    }

    return;
}

int mergeSort(vector<int>& nums, int l, int r){
    if(l>=r) return 0;

    int mid=(r+l)>>1;
    int ans=mergeSort(nums,l,mid)+mergeSort(nums,mid+1,r);

    int left=l;
    int right=mid+1;
    while(left<=mid){
        while(right<=r&&nums[left]>nums[right]){
            right++;
        }
        ans+=(right-mid-1);
        left++;
    }
    merge(nums,l,mid,r);
    return ans;
}

int fun3(vector<int> &nums){
    return mergeSort(nums,0,nums.size()-1);
}