315. 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].

BST优化后面补充，留坑。

public class Solution {    public List<Integer> countSmaller(int[] nums) {        Integer[] res = new Integer[nums.length];        List<Integer> al  = new ArrayList<Integer>();        for (int i = nums.length-1; i >= 0; i--) {            res[i] = find(al, nums[i]);            al.add(res[i], nums[i]);        }        return Arrays.asList(res);    }    public int find(List<Integer> al, int target) {        if(al.size() == 0) return 0;        int start = 0;        int end = al.size()-1;        if(al.get(start) >= target) return 0;        if(al.get(end) < target) return end+1;        while (start < end) {            int mid = (start + end) / 2;            if(al.get(mid) >= target) end = mid;            else start = mid+1;        }        return start;    }}