# [LeetCode] 315. Count of Smaller Numbers After Self

## Problem

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:

Input: [5,2,6,1]
Output: [2,1,1,0]
Explanation:
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.

## Solution

``````class Solution {
public List<Integer> countSmaller(int[] nums) {
List<Integer> res = new ArrayList<>();
if (nums == null || nums.length == 0) return res;
int len = nums.length;
TreeNode root = new TreeNode(nums[len-1]);
for (int i = len-2; i >= 0; i--) {
int count = insertNode(root, nums[i]);
}
Collections.reverse(res);
return res;
}

private int insertNode(TreeNode root, int num) {
int count = 0; //the # of smaller numbers of num
while (root != null) {
if (num <= root.val) {
root.count++;
if (root.left == null) {
root.left = new TreeNode(num);
break;
} else root = root.left;
} else {
count += root.count;
if (root.right == null) {
root.right = new TreeNode(num);
break;
} else root = root.right;
}
}
return count;
}
}
class TreeNode {
int val;
int count = 1;
TreeNode left;
TreeNode right;
public TreeNode(int val) {
this.val = val;
}
}
``````

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