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].

根据题意可以得知，对于输入数组nums，返回数组arrs，其中arrs[i]等于数组nums中位于nums[i]右边且小于nums[i]的元素数目。只需要保存与某个元素的比较结果，很容易想到二分查找树，所以，从后向前遍历数组nums，将元素nums[i]不断插入到二叉树中，同时统计当前小于nums[i]的元素数目。需要注意数组处理nums中的重复元素，由于统计小于nums[i]的元素，因此对于相等的元素也插入到二叉树节点右边，便于计算。代码如下

class BinarySearchTreeNode{BinarySearchTreeNode leftChildren;BinarySearchTreeNode rightChildren;int value;int num;public BinarySearchTreeNode(int v){this.value = v;num = 1;}}public int insert(int v, BinarySearchTreeNode node){int re = 0;while(true){if(v < node.value){node.num ++;if(node.leftChildren == null){node.leftChildren = new BinarySearchTreeNode(v);break;}node = node.leftChildren;}else{node.num ++;// v >= node.value,加上左子树以及node本身re += getNum(node.leftChildren);if(node.value != v)re++;if(node.rightChildren == null){node.rightChildren = new BinarySearchTreeNode(v);break;}node = node.rightChildren;}}return re;}private int getNum(BinarySearchTreeNode node){if(node == null)return 0;elsereturn node.num;}public List<Integer> countSmaller(int[] nums) {        LinkedList<Integer> reList = new LinkedList<Integer>();        if(nums == null || nums.length == 0)        return reList;        reList.addFirst(0);        BinarySearchTreeNode root = new BinarySearchTreeNode(nums[nums.length-1]);        for(int i=nums.length-2; i >= 0; i--){        int el = nums[i];        int re = insert(el,root);        reList.addFirst(re);        }                return reList;}