Count of Smaller Numbers After Self

这串递归代码非常烧脑，先贴出来，慢慢研究，求逆序数，还可以用变形的mergesort：

    class Node {        int key;        int smaller;        int duplicates;        Node left, right;        public Node(int key, int smaller) {            this.key = key;            this.smaller = smaller;            this.duplicates = 1;        }    }    private Node put(Node root, int key, Integer[] answer, int index, int smaller) {        if (root == null) {            answer[index] = smaller;            return new Node(key, 0);        }        if (key < root.key) {            root.smaller++;            root.left = put(root.left, key, answer, index, smaller);        } else if (key > root.key) {            root.right = put(root.right, key, answer, index, smaller + root.smaller + root.duplicates);        } else {            root.duplicates++;            answer[index] = smaller + root.smaller;        }        return root;    }    public List<Integer> countSmaller(int[] nums) {        int n = nums.length;        Integer[] answer = new Integer[n];        Node root = null;        for (int i = n - 1; i >= 0; i--) {            root = put(root, nums[i], answer, i, 0);        }        return Arrays.asList(answer);    }