[leetcode] Count of Smaller Numbers After Self

題目：

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:

Given nums = [5, 2, 6, 1]

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.
Return the array [2, 1, 1, 0].

分析：很顯然不能用時間復雜度為O(N²)的暴力窮舉法，下面先給出一個運行時間為102ms的AC答案：

public class Solution {
    private void add(int[] bit, int i, int val) {
        for (; i < bit.length; i += 1) bit[i] += val;
    }

    private int query(int[] bit, int i) {
        return bit[i];
    }

    public List<Integer> countSmaller(int[] nums) {
        int[] tmp = nums.clone();
        Arrays.sort(tmp);
        for (int i = 0; i < nums.length; i++) {
            nums[i] = Arrays.binarySearch(tmp, nums[i]);
        }
        int[] bit = new int[nums.length];
        Integer[] ans = new Integer[nums.length];
        for (int i = nums.length - 1; i >= 0; i--) {
            ans[i] = query(bit, nums[i]);
            add(bit, nums[i]+1, 1);
        }
        return Arrays.asList(ans);
    }
}

上面的答案雖然AC了，但若原始海量數據本身是降序排列時，再運行時間也為O(N²)，下面給出運用樹狀數組(Binary Indexed Tree (Fenwick tree))的解法，使時間復雜度將為O(NlgN)，樹狀數組的相關知識可參考博客樹狀數組.

    private int lowbit(int x) {
        return x&(-x);
    }
    
    private void add(int[] bit, int i, int val) {
        for (; i < bit.length; i += lowbit(i)) bit[i] += val;
    }

    private int query(int[] bit, int i) {
        int ans = 0;
        for (; i > 0; i -= lowbit(i)) ans += bit[i];
        return ans;
    }

    public List<Integer> countSmaller(int[] nums) {
        int[] tmp = nums.clone();
        Arrays.sort(tmp);
        for (int i = 0; i < nums.length; i++) nums[i] = Arrays.binarySearch(tmp, nums[i]);
        int[] bit = new int[nums.length];
        Integer[] ans = new Integer[nums.length];
        for (int i = nums.length - 1; i >= 0; i--) {
            ans[i] = query(bit, nums[i]);
            add(bit, nums[i]+1, 1);
        }
        return Arrays.asList(ans);
    }

最后給出線段樹(segment tree)的解法，時間復雜度也為O(NlgN)，相似題目請參考Range Sum Query - Mutable，代碼如下：

class SegmentTreeNode {
    int start, end;
    int num;
    SegmentTreeNode ltree, rtree;
    public SegmentTreeNode(int s, int e) {
        start = s;
        end = e;
    }
}

public class Solution { 
    SegmentTreeNode root = null;
    
    public SegmentTreeNode buildTree(int[] nums, int left, int right) {
        SegmentTreeNode root = new SegmentTreeNode(left, right);
        if (left != right) {
            int mid = left + (right - left)/2;
            root.ltree = buildTree(nums, left, mid);
            root.rtree = buildTree(nums, mid+1, right);
        }
        return root;
    }
    
    private void update(SegmentTreeNode root, int i, int val) {
        if (root.start == root.end) {
            root.num += 1;
        } else {
            int mid = root.start + (root.end - root.start)/2;
            if (i <= mid) {
                update(root.ltree, i, val);
            } else {
                update(root.rtree, i, val);
            }
            root.num = root.ltree.num + root.rtree.num;
        }
    }

    private int query(SegmentTreeNode root, int i, int j) {
        if (root.start == i && root.end == j) {
            return root.num;
        } else {
            int mid = root.start + (root.end - root.start)/2;
            if (j <= mid) {
                return query(root.ltree, i, j);
            } else if (i > mid) {
                return query(root.rtree, i, j);
            } else {
                return query(root.ltree, i, root.ltree.end) + query(root.rtree, root.rtree.start, j);
            }
        }
    }

    public List<Integer> countSmaller(int[] nums) {
        int[] tmp = nums.clone();
        Arrays.sort(tmp);
        for (int i = 0; i < nums.length; i++) {
            nums[i] = Arrays.binarySearch(tmp, nums[i]) + 1;
        }
        int[] bit = new int[nums.length + 1];
        root = buildTree(bit, 0, bit.length-1);
        Integer[] ans = new Integer[nums.length];
        for (int i = nums.length - 1; i >= 0; i--) {
            ans[i] = query(root, 0, nums[i] - 1);
            update(root, nums[i], 1);
        }
        return Arrays.asList(ans);
    }
}