---
comments: true
difficulty: 困难
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0315.Count%20of%20Smaller%20Numbers%20After%20Self/README.md
tags:
    - 树状数组
    - 线段树
    - 数组
    - 二分查找
    - 分治
    - 有序集合
    - 归并排序
---

<!-- problem:start -->

# [315. 计算右侧小于当前元素的个数](https://leetcode.cn/problems/count-of-smaller-numbers-after-self)

[English Version](/solution/0300-0399/0315.Count%20of%20Smaller%20Numbers%20After%20Self/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给你一个整数数组 <code>nums</code><em> </em>，按要求返回一个新数组&nbsp;<code>counts</code><em> </em>。数组 <code>counts</code> 有该性质： <code>counts[i]</code> 的值是&nbsp; <code>nums[i]</code> 右侧小于&nbsp;<code>nums[i]</code> 的元素的数量。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>nums = [5,2,6,1]
<strong>输出：</strong><code>[2,1,1,0] 
<strong>解释：</strong></code>
5 的右侧有 <strong>2 </strong>个更小的元素 (2 和 1)
2 的右侧仅有 <strong>1 </strong>个更小的元素 (1)
6 的右侧有 <strong>1 </strong>个更小的元素 (1)
1 的右侧有 <strong>0 </strong>个更小的元素
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>nums = [-1]
<strong>输出：</strong>[0]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>nums = [-1,-1]
<strong>输出：</strong>[0,0]
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：树状数组

树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：

1. **单点更新** `update(x, delta)`： 把序列 x 位置的数加上一个值 delta；
1. **前缀和查询** `query(x)`：查询序列 `[1,...x]` 区间的区间和，即位置 x 的前缀和。

这两个操作的时间复杂度均为 $O(\log n)$。

树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。

比如给定数组 `a[5] = {2, 5, 3, 4, 1}`，求 `b[i] = 位置 i 左边小于等于 a[i] 的数的个数`。对于此例，`b[5] = {0, 1, 1, 2, 0}`。

解决方案是直接遍历数组，每个位置先求出 `query(a[i])`，然后再修改树状数组 `update(a[i], 1)` 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。

<!-- tabs:start -->

#### Python3

```python
class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x > 0:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        alls = sorted(set(nums))
        m = {v: i for i, v in enumerate(alls, 1)}
        tree = BinaryIndexedTree(len(m))
        ans = []
        for v in nums[::-1]:
            x = m[v]
            tree.update(x, 1)
            ans.append(tree.query(x - 1))
        return ans[::-1]
```

#### Java

```java
class Solution {
    public List<Integer> countSmaller(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int v : nums) {
            s.add(v);
        }
        List<Integer> alls = new ArrayList<>(s);
        alls.sort(Comparator.comparingInt(a -> a));
        int n = alls.size();
        Map<Integer, Integer> m = new HashMap<>(n);
        for (int i = 0; i < n; ++i) {
            m.put(alls.get(i), i + 1);
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        LinkedList<Integer> ans = new LinkedList<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            int x = m.get(nums[i]);
            tree.update(x, 1);
            ans.addFirst(tree.query(x - 1));
        }
        return ans;
    }
}

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}
```

#### C++

```cpp
class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1) {}

    void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        vector<int> alls(s.begin(), s.end());
        sort(alls.begin(), alls.end());
        unordered_map<int, int> m;
        int n = alls.size();
        for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
        BinaryIndexedTree* tree = new BinaryIndexedTree(n);
        vector<int> ans(nums.size());
        for (int i = nums.size() - 1; i >= 0; --i) {
            int x = m[nums[i]];
            tree->update(x, 1);
            ans[i] = tree->query(x - 1);
        }
        return ans;
    }
};
```

#### Go

```go
type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= this.lowbit(x)
	}
	return s
}

func countSmaller(nums []int) []int {
	s := make(map[int]bool)
	for _, v := range nums {
		s[v] = true
	}
	var alls []int
	for v := range s {
		alls = append(alls, v)
	}
	sort.Ints(alls)
	m := make(map[int]int)
	for i, v := range alls {
		m[v] = i + 1
	}
	ans := make([]int, len(nums))
	tree := newBinaryIndexedTree(len(alls))
	for i := len(nums) - 1; i >= 0; i-- {
		x := m[nums[i]]
		tree.update(x, 1)
		ans[i] = tree.query(x - 1)
	}
	return ans
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二：线段树

线段树将整个区间分割为多个不连续的子区间，子区间的数量不超过 `log(width)`。更新某个元素的值，只需要更新 `log(width)` 个区间，并且这些区间都包含在一个包含该元素的大区间内。

-   线段树的每个节点代表一个区间；
-   线段树具有唯一的根节点，代表的区间是整个统计范围，如 `[1, N]`；
-   线段树的每个叶子节点代表一个长度为 1 的元区间 `[x, x]`；
-   对于每个内部节点 `[l, r]`，它的左儿子是 `[l, mid]`，右儿子是 `[mid + 1, r]`, 其中 `mid = ⌊(l + r) / 2⌋` (即向下取整)。

<!-- tabs:start -->

#### Python3

```python
class Node:
    def __init__(self):
        self.l = 0
        self.r = 0
        self.v = 0


class SegmentTree:
    def __init__(self, n):
        self.tr = [Node() for _ in range(n << 2)]
        self.build(1, 1, n)

    def build(self, u, l, r):
        self.tr[u].l = l
        self.tr[u].r = r
        if l == r:
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)

    def modify(self, u, x, v):
        if self.tr[u].l == x and self.tr[u].r == x:
            self.tr[u].v += v
            return
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        if x <= mid:
            self.modify(u << 1, x, v)
        else:
            self.modify(u << 1 | 1, x, v)
        self.pushup(u)

    def query(self, u, l, r):
        if self.tr[u].l >= l and self.tr[u].r <= r:
            return self.tr[u].v
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        v = 0
        if l <= mid:
            v += self.query(u << 1, l, r)
        if r > mid:
            v += self.query(u << 1 | 1, l, r)
        return v

    def pushup(self, u):
        self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v


class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        s = sorted(set(nums))
        m = {v: i for i, v in enumerate(s, 1)}
        tree = SegmentTree(len(s))
        ans = []
        for v in nums[::-1]:
            x = m[v]
            ans.append(tree.query(1, 1, x - 1))
            tree.modify(1, x, 1)
        return ans[::-1]
```

#### Java

```java
class Solution {
    public List<Integer> countSmaller(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int v : nums) {
            s.add(v);
        }
        List<Integer> alls = new ArrayList<>(s);
        alls.sort(Comparator.comparingInt(a -> a));
        int n = alls.size();
        Map<Integer, Integer> m = new HashMap<>(n);
        for (int i = 0; i < n; ++i) {
            m.put(alls.get(i), i + 1);
        }
        SegmentTree tree = new SegmentTree(n);
        LinkedList<Integer> ans = new LinkedList<>();
        for (int i = nums.length - 1; i >= 0; --i) {
            int x = m.get(nums[i]);
            tree.modify(1, x, 1);
            ans.addFirst(tree.query(1, 1, x - 1));
        }
        return ans;
    }
}

class Node {
    int l;
    int r;
    int v;
}

class SegmentTree {
    private Node[] tr;

    public SegmentTree(int n) {
        tr = new Node[4 * n];
        for (int i = 0; i < tr.length; ++i) {
            tr[i] = new Node();
        }
        build(1, 1, n);
    }

    public void build(int u, int l, int r) {
        tr[u].l = l;
        tr[u].r = r;
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }

    public void modify(int u, int x, int v) {
        if (tr[u].l == x && tr[u].r == x) {
            tr[u].v += v;
            return;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (x <= mid) {
            modify(u << 1, x, v);
        } else {
            modify(u << 1 | 1, x, v);
        }
        pushup(u);
    }

    public void pushup(int u) {
        tr[u].v = tr[u << 1].v + tr[u << 1 | 1].v;
    }

    public int query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return tr[u].v;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        int v = 0;
        if (l <= mid) {
            v += query(u << 1, l, r);
        }
        if (r > mid) {
            v += query(u << 1 | 1, l, r);
        }
        return v;
    }
}
```

#### C++

```cpp
class Node {
public:
    int l;
    int r;
    int v;
};

class SegmentTree {
public:
    vector<Node*> tr;

    SegmentTree(int n) {
        tr.resize(4 * n);
        for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
        build(1, 1, n);
    }

    void build(int u, int l, int r) {
        tr[u]->l = l;
        tr[u]->r = r;
        if (l == r) return;
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }

    void modify(int u, int x, int v) {
        if (tr[u]->l == x && tr[u]->r == x) {
            tr[u]->v += v;
            return;
        }
        int mid = (tr[u]->l + tr[u]->r) >> 1;
        if (x <= mid)
            modify(u << 1, x, v);
        else
            modify(u << 1 | 1, x, v);
        pushup(u);
    }

    void pushup(int u) {
        tr[u]->v = tr[u << 1]->v + tr[u << 1 | 1]->v;
    }

    int query(int u, int l, int r) {
        if (tr[u]->l >= l && tr[u]->r <= r) return tr[u]->v;
        int mid = (tr[u]->l + tr[u]->r) >> 1;
        int v = 0;
        if (l <= mid) v += query(u << 1, l, r);
        if (r > mid) v += query(u << 1 | 1, l, r);
        return v;
    }
};

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        unordered_set<int> s(nums.begin(), nums.end());
        vector<int> alls(s.begin(), s.end());
        sort(alls.begin(), alls.end());
        unordered_map<int, int> m;
        int n = alls.size();
        for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
        SegmentTree* tree = new SegmentTree(n);
        vector<int> ans(nums.size());
        for (int i = nums.size() - 1; i >= 0; --i) {
            int x = m[nums[i]];
            tree->modify(1, x, 1);
            ans[i] = tree->query(1, 1, x - 1);
        }
        return ans;
    }
};
```

#### Go

```go
type Pair struct {
	val   int
	index int
}

var (
	tmp   []Pair
	count []int
)

func countSmaller(nums []int) []int {
	tmp, count = make([]Pair, len(nums)), make([]int, len(nums))
	array := make([]Pair, len(nums))
	for i, v := range nums {
		array[i] = Pair{val: v, index: i}
	}
	sorted(array, 0, len(array)-1)
	return count
}

func sorted(arr []Pair, low, high int) {
	if low >= high {
		return
	}
	mid := low + (high-low)/2
	sorted(arr, low, mid)
	sorted(arr, mid+1, high)
	merge(arr, low, mid, high)
}

func merge(arr []Pair, low, mid, high int) {
	left, right := low, mid+1
	idx := low
	for left <= mid && right <= high {
		if arr[left].val <= arr[right].val {
			count[arr[left].index] += right - mid - 1
			tmp[idx], left = arr[left], left+1
		} else {
			tmp[idx], right = arr[right], right+1
		}
		idx++
	}
	for left <= mid {
		count[arr[left].index] += right - mid - 1
		tmp[idx] = arr[left]
		idx, left = idx+1, left+1
	}
	for right <= high {
		tmp[idx] = arr[right]
		idx, right = idx+1, right+1
	}
	// 排序
	for i := low; i <= high; i++ {
		arr[i] = tmp[i]
	}
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法三：归并排序

<!-- solution:end -->

<!-- problem:end -->