Question

493. 翻转对

English Version

题目描述

给定一个数组 nums ，如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个_重要翻转对_。

你需要返回给定数组中的重要翻转对的数量。

示例 1:

输入: [1,3,2,3,1]
输出: 2

示例 2:

输入: [2,4,3,5,1]
输出: 3

注意:

1. 给定数组的长度不会超过50000。
2. 输入数组中的所有数字都在32位整数的表示范围内。

解法

方法一：归并排序

归并排序的过程中，如果左边的数大于右边的数，则右边的数与左边的数之后的数都构成逆序对。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组长度。

class Solution:
  def reversePairs(self, nums: List[int]) -> int:
    def merge_sort(l, r):
      if l >= r:
        return 0
      mid = (l + r) >> 1
      ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
      t = []
      i, j = l, mid + 1
      while i <= mid and j <= r:
        if nums[i] <= 2 * nums[j]:
          i += 1
        else:
          ans += mid - i + 1
          j += 1
      i, j = l, mid + 1
      while i <= mid and j <= r:
        if nums[i] <= nums[j]:
          t.append(nums[i])
          i += 1
        else:
          t.append(nums[j])
          j += 1
      t.extend(nums[i : mid + 1])
      t.extend(nums[j : r + 1])
      nums[l : r + 1] = t
      return ans

    return merge_sort(0, len(nums) - 1)

class Solution {
  private int[] nums;
  private int[] t;

  public int reversePairs(int[] nums) {
    this.nums = nums;
    int n = nums.length;
    this.t = new int[n];
    return mergeSort(0, n - 1);
  }

  private int mergeSort(int l, int r) {
    if (l >= r) {
      return 0;
    }
    int mid = (l + r) >> 1;
    int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
    int i = l, j = mid + 1, k = 0;
    while (i <= mid && j <= r) {
      if (nums[i] <= nums[j] * 2L) {
        ++i;
      } else {
        ans += mid - i + 1;
        ++j;
      }
    }
    i = l;
    j = mid + 1;
    while (i <= mid && j <= r) {
      if (nums[i] <= nums[j]) {
        t[k++] = nums[i++];
      } else {
        t[k++] = nums[j++];
      }
    }
    while (i <= mid) {
      t[k++] = nums[i++];
    }
    while (j <= r) {
      t[k++] = nums[j++];
    }
    for (i = l; i <= r; ++i) {
      nums[i] = t[i - l];
    }
    return ans;
  }
}

class Solution {
public:
  int reversePairs(vector<int>& nums) {
    int n = nums.size();
    int t[n];
    function<int(int, int)> mergeSort = [&](int l, int r) -> int {
      if (l >= r) {
        return 0;
      }
      int mid = (l + r) >> 1;
      int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
      int i = l, j = mid + 1, k = 0;
      while (i <= mid && j <= r) {
        if (nums[i] <= nums[j] * 2LL) {
          ++i;
        } else {
          ans += mid - i + 1;
          ++j;
        }
      }
      i = l;
      j = mid + 1;
      while (i <= mid && j <= r) {
        if (nums[i] <= nums[j]) {
          t[k++] = nums[i++];
        } else {
          t[k++] = nums[j++];
        }
      }
      while (i <= mid) {
        t[k++] = nums[i++];
      }
      while (j <= r) {
        t[k++] = nums[j++];
      }
      for (i = l; i <= r; ++i) {
        nums[i] = t[i - l];
      }
      return ans;
    };
    return mergeSort(0, n - 1);
  }
};

func reversePairs(nums []int) int {
  n := len(nums)
  t := make([]int, n)
  var mergeSort func(l, r int) int
  mergeSort = func(l, r int) int {
    if l >= r {
      return 0
    }
    mid := (l + r) >> 1
    ans := mergeSort(l, mid) + mergeSort(mid+1, r)
    i, j, k := l, mid+1, 0
    for i <= mid && j <= r {
      if nums[i] <= nums[j]*2 {
        i++
      } else {
        ans += mid - i + 1
        j++
      }
    }
    i, j = l, mid+1
    for i <= mid && j <= r {
      if nums[i] <= nums[j] {
        t[k] = nums[i]
        k, i = k+1, i+1
      } else {
        t[k] = nums[j]
        k, j = k+1, j+1
      }
    }
    for ; i <= mid; i, k = i+1, k+1 {
      t[k] = nums[i]
    }
    for ; j <= r; j, k = j+1, k+1 {
      t[k] = nums[j]
    }
    for i = l; i <= r; i++ {
      nums[i] = t[i-l]
    }
    return ans
  }
  return mergeSort(0, n-1)
}

方法二：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 reversePairs(self, nums: List[int]) -> int:
    s = set()
    for num in nums:
      s.add(num)
      s.add(num * 2)
    alls = sorted(s)
    m = {v: i for i, v in enumerate(alls, 1)}
    ans = 0
    tree = BinaryIndexedTree(len(m))
    for num in nums[::-1]:
      ans += tree.query(m[num] - 1)
      tree.update(m[num * 2], 1)
    return ans

class Solution {
  public int reversePairs(int[] nums) {
    TreeSet<Long> ts = new TreeSet<>();
    for (int num : nums) {
      ts.add((long) num);
      ts.add((long) num * 2);
    }
    Map<Long, Integer> m = new HashMap<>();
    int idx = 0;
    for (long num : ts) {
      m.put(num, ++idx);
    }
    BinaryIndexedTree tree = new BinaryIndexedTree(m.size());
    int ans = 0;
    for (int i = nums.length - 1; i >= 0; --i) {
      int x = m.get((long) nums[i]);
      ans += tree.query(x - 1);
      tree.update(m.get((long) nums[i] * 2), 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:
  int reversePairs(vector<int>& nums) {
    set<long long> s;
    for (int num : nums) {
      s.insert(num);
      s.insert(num * 2ll);
    }
    unordered_map<long long, int> m;
    int idx = 0;
    for (long long num : s) m[num] = ++idx;
    BinaryIndexedTree* tree = new BinaryIndexedTree(m.size());
    int ans = 0;
    for (int i = nums.size() - 1; i >= 0; --i) {
      ans += tree->query(m[nums[i]] - 1);
      tree->update(m[nums[i] * 2ll], 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 reversePairs(nums []int) int {
  s := make(map[int]bool)
  for _, num := range nums {
    s[num] = true
    s[num*2] = true
  }
  var alls []int
  for num := range s {
    alls = append(alls, num)
  }
  sort.Ints(alls)
  m := make(map[int]int)
  for i, num := range alls {
    m[num] = i + 1
  }
  tree := newBinaryIndexedTree(len(m))
  ans := 0
  for i := len(nums) - 1; i >= 0; i-- {
    ans += tree.query(m[nums[i]] - 1)
    tree.update(m[nums[i]*2], 1)
  }
  return ans
}

方法三：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(4 * n)]
    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 += 1
      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 pushup(self, u):
    self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v

  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


class Solution:
  def reversePairs(self, nums: List[int]) -> int:
    s = set()
    for num in nums:
      s.add(num)
      s.add(num * 2)
    alls = sorted(s)
    m = {v: i for i, v in enumerate(alls, 1)}
    tree = SegmentTree(len(m))
    ans = 0
    for v in nums[::-1]:
      x = m[v]
      ans += tree.query(1, 1, x - 1)
      tree.modify(1, m[v * 2], 1)
    return ans

class Solution {
  public int reversePairs(int[] nums) {
    TreeSet<Long> ts = new TreeSet<>();
    for (int num : nums) {
      ts.add((long) num);
      ts.add((long) num * 2);
    }
    Map<Long, Integer> m = new HashMap<>();
    int idx = 0;
    for (long num : ts) {
      m.put(num, ++idx);
    }
    SegmentTree tree = new SegmentTree(m.size());
    int ans = 0;
    for (int i = nums.length - 1; i >= 0; --i) {
      int x = m.get((long) nums[i]);
      ans += tree.query(1, 1, x - 1);
      tree.modify(1, m.get((long) nums[i] * 2), 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:
  int reversePairs(vector<int>& nums) {
    set<long long> s;
    for (int num : nums) {
      s.insert(num);
      s.insert(num * 2ll);
    }
    unordered_map<long long, int> m;
    int idx = 0;
    for (long long num : s) m[num] = ++idx;
    SegmentTree* tree = new SegmentTree(m.size());
    int ans = 0;
    for (int i = nums.size() - 1; i >= 0; --i) {
      ans += tree->query(1, 1, m[nums[i]] - 1);
      tree->modify(1, m[nums[i] * 2ll], 1);
    }
    return ans;
  }
};