"Leetcode" Count of Smaller Numbers after self

Source: Internet
Author: User

Title Link: https://leetcode.com/problems/count-of-smaller-numbers-after-self/


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] .


2 1 1 0 smaller element.

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


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


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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