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315. Count of Smaller Numbers After Self

2024-09-05 20:51:12   217人阅读

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:

Given nums = [5, 2, 6, 1]

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.

Return the array [2, 1, 1, 0].

解题思路：求逆序数对，可以用线段树或者后缀数组求解。由于数值可能为负数，所以需要离散化或者加上最小的数+1。

离散化：

#define lowbit(i) (i&(-i))
const int maxn = 100010;
struct node{
    int val,pos;
}temp[maxn];
bool cmp(node a,node b){
    return a.val<b.val;
}
class Solution {
public:
    int c[maxn]={0};
    int A[maxn]={0};
    void update(int x,int v){
        for(int i=x;i<maxn;i+=lowbit(i)){
            c[i]+=v;
        }
    }
    int getSum(int x){
        int sum=0;
        for(int i=x;i>0;i-=lowbit(i)){
            sum+=c[i];
        }
        return sum;
    }
    
    vector<int> countSmaller(vector<int>& nums) {
        vector<int>ans;
        int n=nums.size();
        for(int i=0;i<n;i++){
            temp[i].pos=i;
            temp[i].val=nums[i];
        }
        sort(temp,temp+n,cmp);
        for(int i=0;i<n;i++){
            if(i==0||temp[i].val!=temp[i-1].val){
                A[temp[i].pos]=i+1;
            }
            else  A[temp[i].pos]= A[temp[i-1].pos];
        }
        for(int i=n-1;i>=0;i--){
            update(A[i],1);
            ans.push_back(getSum(A[i]-1));
        }
        reverse(ans.begin(),ans.end());
        return ans;
    }
};

 最小数+1：

#define lowbit(i) (i&(-i))
const int maxn = 100010;
class Solution {
public:
    int c[maxn]={0};
    void update(int x,int v){
        for(int i=x;i<maxn;i+=lowbit(i)){
            c[i]+=v;
        }
    }
    int getSum(int x){
        int sum=0;
        for(int i=x;i>0;i-=lowbit(i)){
            sum+=c[i];
        }
        return sum;
    }
    vector<int> countSmaller(vector<int>& nums) {
        vector<int>ans;
        int n=nums.size(),mmin=INT_MAX;
        for(int i=0;i<n;i++){
            mmin=min(mmin,nums[i]);
        }
        for(int i=0;i<n;i++){
            nums[i]=nums[i]-mmin+1;
        }
        for(int i=n-1;i>=0;i--){
            update(nums[i],1);
            ans.push_back(getSum(nums[i]-1));
        }
        reverse(ans.begin(),ans.end());
        return ans;
    }
};

 

315. Count of Smaller Numbers After Self


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