[LeetCode][JavaScript]Count of Smaller Numbers After Self

Posted 2020-06-17

Count of Smaller Numbers After Self

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].

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

 

 

---

 

 

在数组中统计当前数之后比他小的数的个数，复杂度要小于O(n^2)。

首先注意到应该从右往左遍历，记录下访问过的数，这样就转化成了子问题：

怎么从一个数据结构中快速找出所有比他小的数，并且这个数据结构要支持O(logn)的插入操作。

那么就想到用二叉搜索树，所有左子树的数据都比当前节点小，这边用数组实现BST。

二分查找插入的位置，这个位置恰好也代表了左边有几个比他小的数。

 

 1 /**
 2  * @param {number[]} nums
 3  * @return {number[]}
 4  */
 5 var countSmaller = function(nums) {
 6     var i, bst = [], result = [], index, tt1, tt2;
 7     for(i = nums.length - 1; i >= 0; i--){
 8         index = getBSTIndex(bst, nums[i]);
 9         result.unshift(index); 
10         bst.splice(index, 0, nums[i]);
11     }
12     return result;
13 
14     function getBSTIndex(bst, num){
15         if(bst.length === 0 ) return 0;
16         var i = 0, j = bst.length - 1, middle;
17         if(bst[j] < num) return bst.length;
18         if(bst[i] >= num) return 0;
19         while(i + 1 < j){
20             middle = (i + j) >> 1;
21             if(bst[middle] < num)
22                 i = middle + 1;
23             else
24                 j = middle;
25         }
26         if(bst[i] >= num) return i;
27         return j;
28     }
29 };