leetcode - Count of Smaller Numbers After Self

题目：

    https://leetcode.com/problems/count-of-smaller-numbers-after-self/

思路：

    使用线段树，参考这里。

class Solution
{
	class Node//线段树节点
	{
	public:
		int ll,rr,cnt; // 在范围[ll,rr]内的值有cnt个
		Node *left,*right;
		Node(int a,int b):ll(a),rr(b),cnt(0),left(NULL),right(NULL){}
	};
	class Tree//线段树
	{
		Node* buildtree(int begin,int end)
		{
			if(begin==end)
			{
				Node* res=new Node(begin,end);
				return res;
			}
			else
			{
				int mid=begin+(end-begin)/2;
				Node* r1=buildtree(begin,mid),*r2=buildtree(mid+1,end);
				Node* res=new Node(begin,end);
				res->left=r1;
				res->right=r2;
				return res;
			}
		}

		void update_core( Node* root,int val)
		{
			if(root==NULL || valll || val> root->rr)
				return;
			++root->cnt;
			update_core(root->left,val);
			update_core(root->right,val);
		}

		int search_core(Node* root,int  low,int up)
		{
			if(root==NULL  || upll || low>root->rr)
				return 0;
			if(root->ll>=low && root-> rr<=up)
				return root->cnt;
			return search_core(root->left,low,up)+search_core(root->right,low,up);
		}

	public:
		Node *root;
		Tree(int begin,int end)
		{
			root=buildtree(begin,end);
		}

		void update(int val)
		{
			update_core(root,val);
		}

		int search(int low,int up)
		{
			if(low>up)
				return 0;
			return search_core(root,low,up);
		}
	};

	int binarysearch(vector& nums,int n)
	{
		int l=0,r=nums.size()-1;
		while(l<=r)
		{
			int mid=l+(r-l)/2;
			if(nums[mid]==n)
				return mid;
			else if(nums[mid]>n)
				r=mid-1;
			else
				l=mid+1;
		}
		return -1;
	}

public:
	vector countSmaller(vector& nums) 
	{
		if(nums.empty())
			return nums;
		vectortmp=nums;
		sort(tmp.begin(),tmp.end());
		for(int i=0;i res(nums.size(),0);
		//从右向左遍历（因为题目求的是右边比该值小的数的数量)
		for(int i=nums.size()-1;i>=0;--i)
		{
			if(nums[i]!=0) //nums[i]==0，即最小的数，此时不用更新res数组
				res[i]=tree.search(0,nums[i]-1);
			//每次遍历都往线段树压入对应的值
			tree.update(nums[i]);
		}
		return res;
	}
};