leetcode315

 public class Solution {
     public List<Integer> countSmaller(int[] nums) {
         List<Integer> res = new ArrayList<>();
         if(nums == null || nums.length == 0) return res;
         TreeNode root = new TreeNode(nums[nums.length - 1]);
         res.add(0);
         for(int i = nums.length - 2; i >= 0; i--) {
             int count = insertNode(root, nums[i]);
             res.add(count);
         }
         Collections.reverse(res);
         return res;
     }
 
     public int insertNode(TreeNode root, int val) {
         int thisCount = 0;
         while(true) {
             if(val <= root.val) {
                 root.count++;
                 if(root.left == null) {
                     root.left = new TreeNode(val); break;
                 } else {
                     root = root.left;
                 }
             } else {
                 thisCount += root.count;
                 if(root.right == null) {
                     root.right = new TreeNode(val); break;
                 } else {
                     root = root.right;
                 }
             }
         }
         return thisCount;
     }
 }
 
 class TreeNode {
     TreeNode left;
     TreeNode right;
     int val;
     int count = 1;
     public TreeNode(int val) {
         this.val = val;
     }
 }

从后向前进行判断，线性的搜索时间复杂度比较高，因此建立一个二叉搜索树，每次插入节点的时候，更新父节点的“右侧小”的数字的数量。

参考：https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76587/Easiest-Java-solution