LeetCode 315. Count of Smaller Numbers After Self

原題連接在這裏：https://leetcode.com/problems/count-of-smaller-numbers-after-self/

題目：

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].

題解：

從右向左掃描數組nums, 用每一個nums[i]生成TreeNode作BST查找, 返回count 加到 res中.

BST當前節點node左側所有是值小於或者等於當前節點node.val的節點，當前結點node.count是左側子樹包括自身的節點總數.

Time Complexity: O(n^2). BST不必定balance. Space: O(n).

AC Java:

 1 public class Solution {
 2     public List<Integer> countSmaller(int[] nums) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         if(nums == null || nums.length == 0){
 5             return res;
 6         }
 7         res.add(0);
 8         TreeNode root = new TreeNode(nums[nums.length-1]);
 9         for(int i = nums.length-2; i>=0; i--){
10             int curCount = addNode(root, nums[i]);
11             res.add(curCount);
12         }
13         Collections.reverse(res);
14         return res;
15     }
16     
17     private int addNode(TreeNode root, int val){
18         int curCount = 0;
19         while(true){
20             if(root.val >= val){
21                 root.count++;
22                 if(root.left == null){
23                     root.left = new TreeNode(val);
24                     break;
25                 }else{
26                     root = root.left;
27                 }
28             }else{
29                 curCount+=root.count;
30                 if(root.right == null){
31                     root.right = new TreeNode(val);
32                     break;
33                 }else{
34                     root = root.right;
35                 }
36             }
37         }
38         return curCount;
39     }
40 }
41 
42 class TreeNode{
43     int val;
44     int count = 1;
45     TreeNode left;
46     TreeNode right;
47     public TreeNode(int val){
48         this.val = val;
49     }
50 }

相似Reverse Pairs.

Reference: http://www.cnblogs.com/yrbbest/p/5068550.html