315. Count of Smaller Numbers After Self

经典题，三种方法解决

• 树状数组
  将数组转变成rank数组及其频率
  loop from end to beginning

class Solution {
    public List countSmaller(int[] nums) {
        //step1, map each unique element with rank (the smallest element rank 1)
        TreeSet ts = new TreeSet<>();  
        for (int num : nums) ts.add(num);                     //O(nlog n)
        int rank = 1;                                     
        Map rankMap = new HashMap<>();
        for (int num : ts) rankMap.put(num, rank++);          //O(k), k is the # of unique elements
        
        //step2, use binary indexed tree to update and get prefix sum
        List res = new ArrayList<>();
        int[] ranks = new int[ts.size() + 1];                 //ranks[i] record the # of element with rank-i
        for (int i = nums.length - 1; i >= 0; i--) {
            int r = rankMap.get(nums[i]);
            update(r, ranks);                                  //O(log k)
            int preSum = getSum(r - 1, ranks);                 //O(log k)
            res.add(preSum);
        }
        Collections.reverse(res);                              //O(nlog n)
        return res;
    }
    private void update(int i, int[] ranks) {
        int diff = 1;
        while (i < ranks.length) {
            ranks[i] += diff;
            i += (i & (-i));
        }
    }
    private int getSum(int i, int[] ranks) {
        int sum = 0;
        while (i > 0) {
            sum += ranks[i];
            i -= (i & (-i));    
        }
        return sum;
    }
}

• 线段树
  相同的思路，都是维持rank数组
  用线段树求prefix sum
  老三样，build，update 和 query

class Solution {
    class TreeNode {
        int begin, end, sum;
        TreeNode left, right;
        public TreeNode(int begin, int end, int sum) {
            this.begin = begin;
            this.end = end;
            this.sum = sum;
        }
    }
    public List countSmaller(int[] nums) {
        //step1, map each unique element with rank (the smallest element rank 1)
        TreeSet ts = new TreeSet<>();  
        for (int num : nums) ts.add(num);                     //O(nlog n)
        int rank = 1;                                     
        Map rankMap = new HashMap<>();
        for (int num : ts) rankMap.put(num, rank++);          //O(k), k is the # of unique elements
        
        //step2, use binary indexed tree to update and get prefix sum
        List res = new ArrayList<>();
        int[] ranks = new int[ts.size() + 1];                 //ranks[i] record the # of element with rank-i
        TreeNode root = build(0, ranks.length-1);             //O(k)
        for (int i = nums.length - 1; i >= 0; i--) {
            int r = rankMap.get(nums[i]);
            ranks[r] += 1;
            update(root, r, ranks[r]);                        //O(log k)
            int preSum = query(root, 0, r - 1);               //O(log k)
            res.add(preSum);
        }
        Collections.reverse(res);                             //O(nlog n)
        return res;
    }
    private TreeNode build(int begin, int end) {
        if (begin == end) {
            return new TreeNode(begin, end, 0);
        }
        int mid = begin + (end - begin) / 2;
        TreeNode root = new TreeNode(begin, end, 0);
        root.left = build(begin, mid);
        root.right = build(mid+1, end);
        return root;
    }
    private void update(TreeNode root, int idx, int val) {
        if (root.begin == root.end && root.begin == idx) {
            root.sum = val;
            return;
        }
        int mid = root.begin + (root.end - root.begin) / 2;
        if (idx <= mid) {
            update(root.left, idx, val);
        } else {
            update(root.right, idx, val);
        }
        root.sum = (root.left == null ? 0 : root.left.sum) + (root.right == null ? 0 : root.right.sum);
    }
    private int query(TreeNode root, int i, int j) {
        if (root.end < i || root.begin > j) {
            return 0;
        }
        if (root.begin >= i && root.end <= j) {
            return root.sum;
        }
        return query(root.left, i, j) + query(root.right, i, j);
    }
}

• 用BST

class Solution {
    class TreeNode {
        int val;
        int dup;                   //当前node的重复数量
        int leftCount;             //leftCount用于计算小于当前node的数量
        TreeNode left, right;
        public TreeNode(int val) {
            this.val = val;
            dup = 0;
            leftCount = 0;
        }
    }
    public List countSmaller(int[] nums) {
        if (nums == null || nums.length == 0) return new ArrayList<>();
        int n = nums.length;
        //build root, and insert
        TreeNode root = new TreeNode(nums[n-1]);
        int[] res = new int[n];
        for (int i = n-1; i >= 0; i--) {
            res[i] = insert(root, nums[i]);
        }
        //change arr to list
        List ans = new ArrayList<>();
        for (int num : res) ans.add(num); 
        return ans;
    }
    private int insert(TreeNode node, int val) {
        int sum = 0;
        while (node.val != val) {
            if (node.val > val) {                                         //小于node的数量加一
                if (node.left == null) node.left = new TreeNode(val);
                node.leftCount++;
                node = node.left;
            } else {
                if (node.right == null) node.right = new TreeNode(val);
                sum += node.leftCount + node.dup;
                node = node.right;
            }
        }
        node.dup++;
        return sum + node.leftCount;
    }
}