315. Count of Smaller Numbers After Self

315. Count of Smaller Numbers After Self

題目連接：https://leetcode.com/problems...

divide and conquer的題，用bst來作，這種求有多少smaller的題通常都是bst。node裏多加一個信息：size表示以node爲subtree的節點數。

public class Solution {
    public List<Integer> countSmaller(int[] nums) {
        /* binary search tree 
         */
        int n = nums.length;
        LinkedList<Integer> res = new LinkedList();
        if(n == 0) return res;
        
        res.add(0);
        Node root = new Node(nums[n-1]);
        for(int i = n - 2; i >= 0; i--) {
            res.addFirst(findSmaller(root, nums[i]));
        }
        return res;
    }
    
    private int findSmaller(Node root, int value) {
        int res = 0;
        while(root != null) {
            root.size += 1;
            if(root.val < value) {
                // add root and all left nodes
                res += 1 + (root.left == null ? 0 : root.left.size);
                if(root.right == null) {
                    root.right = new Node(value);
                    break;
                }
                root = root.right;
            }
            else {
                if(root.left == null) {
                    root.left = new Node(value);
                    break;
                }
                root = root.left;
            }
        }
        return res;
    }
    
    class Node {
        int val;
        Node left;
        Node right;
        // count the size of this subtree
        int size;
        Node(int val) { this.val = val; this.size = 1; }
    }
}

binary index tree也能夠作，由於是統計有多少smaller的，其實就是求從最小值到nums[i] - 1的sum。tree的index是nums[i]，要作一個映射，把nums[i]的值映射到[1, # of unique numbers in nums]之間。因此先把array給sort一下，用一個map來作映射。

public class Solution {
    public List<Integer> countSmaller(int[] nums) {
        /* binary index tree 
         */
        // reflection first, key: nums[i], value: order
        Map<Integer, Integer> map = new HashMap();
        int[] sorted = Arrays.copyOf(nums, nums.length);
        Arrays.sort(sorted);
        // record the order
        int idx = 1;
        for(int i = 0; i < nums.length; i++) {
            if(!map.containsKey(sorted[i])) map.put(sorted[i], idx++);
        }
        // range will be [1, idx]
        BIT t = new BIT(idx);
        LinkedList<Integer> res = new LinkedList();
        for(int i = nums.length - 1; i >= 0; i--) {
            int sum = t.sum(map.get(nums[i]) - 1);
            res.addFirst(t.sum(map.get(nums[i]) - 1));
            t.add(map.get(nums[i]), 1);
        }
        return res;
    }
    
    class BIT {
        int[] tree;
        int n;
        BIT(int n) { this.n = n; tree = new int[n]; }
        // sum the smaller elements
        protected int sum(int i) {
            int res = 0;
            while(i > 0) {
                res += tree[i];
                i -= (i & -i);
            }
            return res;
        }
        
        protected void add(int i, int val) {
            while(i < n) {
                tree[i] += val;
                i += (i & -i);
            }
        }
    }
}

還有merge sort的方法，參考discussion：
https://discuss.leetcode.com/...