leetcode No315. Count of Smaller Numbers After Self

Posted 2023-03-01 Dufre.WC

Question

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:

Input: [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.

Algorithm

Point：Segment Tree

首先构造如下线段树，每个圆圈表示一个结点，结点包含如下信息：

• 线段范围
  • start
  • end
• 线段范围内有几个元素
  • count
• 指向左右结点的指针
  • left（左结点的线段范围是[start, start+(end-start)/2]）
  • right(右结点的线段范围是[start+(end-start)/2+1, end])

如下是一个实例

根据这道题来说，线段树的操作函数有三个：

• build():构造线段树
  • 找出数组中最小和最大的元素，做为根节点的线段范围
  • 比如[2,3,4,1]的线段树根节点范围为[1,4]
• insert(int index)):插入一个元素，并更新线段树的count值
  • 比如实例图中插入元素3，则[1,4],[3,4],[3,3]的结点count值都+1
• count(int start, int end):返回[start, end]的count值
  • 比如要找小于元素(记为x)的个数，就是统计线段树中[min, x-1]的个数，其中min为线段树的最小值

对于这道题来说，步骤如下：

• 遍历数组，找到数组的min和max，构造线段树[min, max]
• 从右往左遍历，记当前元素为x
  • 统计线段树中[min, x-1]范围内的元素，即右侧小于x的元素个数
  • 往线段树中插入元素x，并更新线段树各个结点的count值

Code

struct SegmentTreeNode
        int start;
        int end;
        int count;

        SegmentTreeNode* left;
        SegmentTreeNode* right;

        SegmentTreeNode(int _start, int _end):start(_start),end(_end) 
            count = 0;
            left = NULL;
            right = NULL;
        
;

class Solution 
public:
    SegmentTreeNode* build(int start, int end)
        if (start > end)
            return NULL;
        
        SegmentTreeNode* root = new SegmentTreeNode(start, end);

        if (start == end)
            root->count = 0;
        else
            int mid = start + (end - start)/2;
            root->left = build(start, mid);
            root->right = build(mid+1, end);
        
        return root;
    

    int count(SegmentTreeNode* root, int start, int end)
        if (root == NULL || start>end)
            return 0;
        if (start==root->start && end==root->end)
            return root->count;
        
        int mid = root->start + (root->end - root->start)/2;
        int leftcount = 0, rightcount = 0;

        if (start <= mid)
            if (mid < end)
                leftcount = count(root->left, start, mid);
            else
                leftcount = count(root->left, start, end);
        

        if (mid < end)
            if (start <= mid)
                rightcount = count(root->right, mid+1, end);
            else
                rightcount = count(root->right, start, end);
        

        return (leftcount + rightcount);
    

    void insert(SegmentTreeNode* root, int index, int val)
        if (root->start==index && root->end==index)
            root->count += val;
            return;
        

        int mid = root->start + (root->end - root->start)/2;
        if (index>=root->start && index<=mid)
            insert(root->left, index, val);
        
        if (index>mid && index<=root->end)
            insert(root->right, index, val);
        

        root->count = root->left->count + root->right->count;
    

    vector<int> countSmaller(vector<int>& nums) 
        vector<int> res;
        if (nums.empty())
            return res;
        res.resize(nums.size());
        int start = nums[0];
        int end = nums[0];

        for (int i=1; i<nums.size(); i++)
            start = min(start, nums[i]);
            end = max(end, nums[i]);
        

        SegmentTreeNode* root = build(start, end);

        for (int i=nums.size()-1; i>=0; i--)
            res[i] = count(root, start, nums[i]-1);
            insert(root, nums[i], 1);
          

        return res;      
    
;

欢迎订阅我的公众号Coder101