标签：elements 递归调用 调用 res 增加 treenode ddn private cti
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].
含义：针对一个数组，找到每一位右面第一个小于它的值，没有的话就是0，返回数组
1 public class Solution { 2 private class TreeNode { 3 public int val; 4 public int count = 1; 5 public TreeNode left, right; 6 7 public TreeNode(int val) { 8 this.val = val; 9 } 10 } 11 12 public List<Integer> countSmaller(int[] nums) { 13 // 构造一棵二分搜索树，稍有不同的地方是我们需要加一个变量smaller来记录比当前节点值小的所有节点的个数，我们每插入一个节点，会判断其和根节点的大小， 14 // 如果新的节点值小于根节点值，则其会插入到左子树中，我们此时要增加根节点的smaller，并继续递归调用左子节点的insert。 15 // 如果节点值大于根节点值，则需要递归调用右子节点的insert并加上根节点的smaller，并加1 16 17 List<Integer> res = new ArrayList<>(); 18 if(nums == null || nums.length == 0) { 19 return res; 20 } 21 TreeNode root = new TreeNode(nums[nums.length - 1]); 22 res.add(0); 23 24 for(int i = nums.length - 2; i >= 0; i--) { 25 int count = addNode(root, nums[i]); 26 res.add(count); 27 } 28 29 Collections.reverse(res); 30 return res; 31 } 32 33 private int addNode(TreeNode root, int val) { 34 int curCount = 0; 35 while(true) { 36 if(val <= root.val) { 37 root.count++; // add the inversion count 38 if(root.left == null) { 39 root.left = new TreeNode(val); 40 break; 41 } else { 42 root = root.left; 43 } 44 } else { 45 curCount += root.count; 46 if(root.right == null) { 47 root.right = new TreeNode(val); 48 break; 49 } else { 50 root = root.right; 51 } 52 } 53 } 54 55 return curCount; 56 }
315. Count of Smaller Numbers After Self
标签：elements 递归调用 调用 res 增加 treenode ddn private cti
原文地址：http://www.cnblogs.com/wzj4858/p/7731644.html
