315. Count of Smaller Numbers After Self(二分或者算法导论中的归并求逆序数对)

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example 1:

Input: nums = [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

Constraints:

• 0 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4

 

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        vector<int> res(nums.size(),0);
        //从右向左，将数组有序插入tmp.利用二分查找确定当前数右边比它小的数的个数
        vector<int> tmp;
        for(int i=nums.size()-1;i>=0;i--){
            int left = 0,right=tmp.size()-1;
            //找第一个大于等于当前数的位置。插入其中
            while(left <= right){
                int mid = left+(right-left)/2;
                if(tmp[mid] < nums[i]) left = mid+1;
                else right = mid-1;
            }
            //最后返回的位置是left
            res[i]=left;
            //插入nums[i]
            tmp.insert(tmp.begin()+left,nums[i]);
        }
        return res;
    }
};

//归并:先引入逆序数；不同于逆序数对:

res[nums[i].second] += (j-mid-1);
这个里面坑比较多

class Solution {
public:
    //法二：利用归并排序求逆序对数的方法
    //https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/submissions/

    vector<int> countSmaller(vector<int>& nums) {
        int n=nums.size();
        vector<int> res(n,0);
        if(n==0 || n==1) return res;
        vector<pair<int,int>> tmp(n,pair<int,int>{0,0});
        vector<pair<int,int>> idx;
        for(int i=0;i<n;i++){
            idx.push_back(make_pair(nums[i],i));
        }
        mergesort(idx,tmp,0,n-1,res);
        return res;
    }

//merge的过程将left到right有序重新存入nums.归并nums[left,mid],nums[mid+1,right]
  void merge(vector<pair<int,int>>& nums,vector<pair<int,int>>& tmp,int left,int mid,int right,vector<int>& res) {
        int i=left,j=mid+1,k=left;
        for(;i<=mid&&j<=right;){
            if(nums[i].first<=nums[j].first){
                //不同于算整个数组逆序数
                //这里的i不是之前的i。归并后数字的位置被改变了.所以利用pari记录nums[i]原始位置
                //res[i] += (j-mid-1);
                res[nums[i].second] += (j-mid-1);
                tmp[k++] = nums[i++];
            }else{
                tmp[k++] = nums[j++];
            }
        }
        //还有未归并完成的
        while(i<=mid){
            //先计算res
            res[nums[i].second] += (j-mid-1);
            tmp[k++]=nums[i++];
        }
        while(j<=right){
            tmp[k++]=nums[j++];
        }
        //将tmp重新放入nums，那么nums[left,right]即有序了
        for(int i=left;i<=right;i++){
            nums[i] = tmp[i];
        }
        return;
    }
    //归并排序
    void mergesort(vector<pair<int,int>>& nums,vector<pair<int,int>>& tmp,int left,int right,vector<int>& res) {
        if(left < right){
            int mid = left+(right-left)/2;
            mergesort(nums,tmp,left,mid,res);
            mergesort(nums,tmp,mid+1,right,res);
            //合并nums[left,mid] nums[mid+1,right]
            merge(nums,tmp,left,mid,right,res);
        }
        return;
    }

};