leetcode315

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

    public int insertNode(TreeNode root, int val) {
        int thisCount = 0;
        while(true) {
            if(val <= root.val) {
                root.count++;
                if(root.left == null) {
                    root.left = new TreeNode(val); break;
                } else {
                    root = root.left;
                }
            } else {
                thisCount += root.count;
                if(root.right == null) {
                    root.right = new TreeNode(val); break;
                } else {
                    root = root.right;
                }
            }
        }
        return thisCount;
    }
}

class TreeNode {
    TreeNode left; 
    TreeNode right;
    int val;
    int count = 1;
    public TreeNode(int val) {
        this.val = val;
    }
}

从后向前进行判断，线性的搜索时间复杂度比较高，因此建立一个二叉搜索树，每次插入节点的时候，更新父节点的“右侧小”的数字的数量。

参考：https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76587/Easiest-Java-solution