# "Leetcode" Count of Smaller Numbers after self

Source: Internet
Author: User

Topic:

you is given an integer array nums and that you had to return a new counts array. The counts array have the property where `counts[i" `  is the number of smaller elements to the right Of `nums[i] `.

Example:

`2 1 1 0 smaller element.`

Return the array `[2, 1, 1, 0]` .

Ideas:

Set up a binary search tree, in the process of building, recorded in the array to the right of the number of elements smaller than itself. Algorithm complexity O (NLOGN), with dynamic programming complexity of O (n^2).

Where CNT is the number of elements of the same size of the node, used for repeating element judgments ... can be omitted in fact. Too lazy to change

Algorithm:

`Public list<integer> Countsmaller (int[] nums) {list<integer> List = new arraylist<integer> ();          int res[] = new Int[nums.length];          for (int i = nums.length-1; I >= 0; i--) {Res[i] = insert (nums[i]);          } for (int i:res) {list.add (i);      } return list;        } TreeNode Troot;          Private Integer Insert (int val) {int cnt = 0;              if (troot = = null) {troot = new TreeNode (val);          return CNT;          } TreeNode root = Troot;                  while (root = null) {if (Val < root.val) {root.leftcnt++;                      if (Root.left = = null) {Root.left = new TreeNode (val);                  Break              } else root = Root.left;                  } else if (val > Root.val) {cnt + = root.leftcnt + root.cnt; if (ROOT.Right = = null) {root.right = new TreeNode (val);                  Break              } else root = Root.right;                  } else {cnt + = root.leftcnt;                  root.cnt++;              Break      }} return CNT;   }`

"Leetcode" Count of Smaller Numbers after self

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