Question

315. 计算右侧小于当前元素的个数

English Version

题目描述

给你一个整数数组 nums_ _，按要求返回一个新数组 counts_ _。数组 counts 有该性质： counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。

 

示例 1：

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

示例 2：

输入：nums = [-1]
输出：[0]

示例 3：

输入：nums = [-1,-1]
输出：[0,0]

 

提示：

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104

解法

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；
2. 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。

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]

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;
  }
}

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;
  }
};

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
}

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)。

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]

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;
  }
}

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;
  }
};

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]
  }
}

方法三：归并排序