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

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/

 

 

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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。

 

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var countSmaller = function(nums) {
    var i, bst = [], result = [], index;
    for(i = nums.length - 1; i >= 0; i--){
        index = getBSTIndex(bst, nums[i]);
        result.unshift(index); 
        bst.splice(index, 0, nums[i]);
    }
    return result;

    function getBSTIndex(bst, num){
        if(bst.length === 0) return 0;
        var start = 0, end = bst.length , middle;
        if(bst[start] >= num) return 0;
        if(bst[end] <= num) return bst.length;
        while(start < end){
            middle = (start + end) >> 1;
            if(bst[middle] >= num)
                end = middle;
            else
                start = middle + 1;
        }
        return start;
    }
};