leetcode:Count of Smaller Numbers After Self

最新推荐文章于 2020-05-20 23:06:37 发布

martin_liang

最新推荐文章于 2020-05-20 23:06:37 发布

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本文介绍了一种使用树结构的方法来计算给定整数数组中每个元素右侧有多少个较小的元素，提供了算法实现和详细解释。

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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].

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struct TreeNodei {
    int val;
    int count;
        
    struct TreeNodei *left;
    struct TreeNodei *right;
        
    TreeNodei(int x) : val(x), count(1), left(NULL), right(NULL) 
    //count 初始化为1是表示以这个Node为根节点（包括它自己全部左子树的节点数）
};

class Solution {
    
private:
    int insertNode(TreeNodei *root, TreeNodei* curNode) {
     
     int count = 0;//count表示curNode从最右端到它所在的位置比它小的数目
                   //和node->count是不同的概念
     
     while (true)
     {
        if (root->val>=curNode->val)
        {
            root->count++;
            if (root->left)
            {
                root = root->left;
            }
            else
            {
                root->left = curNode;
                break;
            }
        }
        else
        {
            count += root->count;
            
            if (root->right)
            {
                root = root->right;
            }
            else
            {
                root->right = curNode;
                break;
            }
        }
     }
     
     return count;
}
    
public:
    vector<int> countSmaller(vector<int>& nums) {
        
        vector<int> retVtr(nums.size());
            
        if (nums.size() == 0)
            return retVtr;
            
        TreeNodei *root = new TreeNodei(nums[nums.size()-1]);
        retVtr[nums.size()-1] = 0;
            
        for (int i=nums.size()-2; i>=0; i--)
        {
            TreeNodei* curNode= new TreeNodei(nums[i]);
            int val = insertNode(root, curNode);
            retVtr[i] = val;
        }
        
        return retVtr;
    }
};