LeetCode 315. Count of Smaller Numbers After Self

原题链接在这里：https://leetcode.com/problems/count-of-smaller-numbers-after-self/

题目：

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].

题解：

从右向左扫描数组nums, try to find the position of nums[i] in BST.

For each BST node, it contains its left subtree size count, its duplicate count.

When inserting a new node, returns the sum of smaller count.

Time Complexity: O(n^2). BST不一定balance. Space: O(n).

AC Java:

 class Solution {
     public List<Integer> countSmaller(int[] nums) {
         LinkedList<Integer> res = new LinkedList<>();
         if(nums == null || nums.length == 0){
             return res;
         }

         int n = nums.length;
         res.offerFirst(0);
         TreeNode root = new TreeNode(nums[n - 1]);
         root.count = 1;

         for(int i = n - 2; i >= 0; i--){
             int smallerCount = insert(root, nums[i]);
             res.offerFirst(smallerCount);
         }

         return res;
     }

     private int insert(TreeNode root, int num){
         int smallerCountSum = 0;
         while(root.val != num){
             if(root.val > num){
                 root.leftCount++;
                 if(root.left == null){
                     root.left = new TreeNode(num);
                 }

                 root = root.left;
             }else{
                 smallerCountSum += root.leftCount + root.count;
                 if(root.right == null){
                     root.right = new TreeNode(num);
                 }

                 root = root.right;
             }
         }

         root.count++;
         return smallerCountSum + root.leftCount;
     }
 }

 class TreeNode{
     int val;
     int count;
     int leftCount;
     TreeNode left;
     TreeNode right;

     public TreeNode(int val){
         this.val = val;
         this.count = 0;
         this.leftCount = 0;
     }
 }

类似Reverse Pairs.