Question

面试题 17.08. 马戏团人塔

English Version

题目描述

有个马戏团正在设计叠罗汉的表演节目，一个人要站在另一人的肩膀上。出于实际和美观的考虑，在上面的人要比下面的人矮一点且轻一点。已知马戏团每个人的身高和体重，请编写代码计算叠罗汉最多能叠几个人。

示例：

输入：height = [65,70,56,75,60,68] weight = [100,150,90,190,95,110]
输出：6
解释：从上往下数，叠罗汉最多能叠 6 层：(56,90), (60,95), (65,100), (68,110), (70,150), (75,190)

提示：

• height.length == weight.length <= 10000

解法

方法一：排序 + 离散化 + 树状数组

我们现将所有人按照身高从小到大排序，若身高相同，则按照体重从大到小排序。这样我们可以将问题转换为求体重数组的最长递增子序列的问题。

最长递增子序列的问题可以使用动态规划求解，时间复杂度 $O(n^2)$。但是我们可以使用树状数组来优化求解过程，时间复杂度 $O(n \log n)$。

空间复杂度 $O(n)$。其中 $n$ 为人数。

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

  def update(self, x, delta):
    while x <= self.n:
      self.c[x] = max(self.c[x], delta)
      x += x & -x

  def query(self, x):
    s = 0
    while x:
      s = max(s, self.c[x])
      x -= x & -x
    return s


class Solution:
  def bestSeqAtIndex(self, height: List[int], weight: List[int]) -> int:
    arr = list(zip(height, weight))
    arr.sort(key=lambda x: (x[0], -x[1]))
    alls = sorted({w for _, w in arr})
    m = {w: i for i, w in enumerate(alls, 1)}
    tree = BinaryIndexedTree(len(m))
    ans = 1
    for _, w in arr:
      x = m[w]
      t = tree.query(x - 1) + 1
      ans = max(ans, t)
      tree.update(x, t)
    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 val) {
    while (x <= n) {
      this.c[x] = Math.max(this.c[x], val);
      x += x & -x;
    }
  }

  public int query(int x) {
    int s = 0;
    while (x > 0) {
      s = Math.max(s, this.c[x]);
      x -= x & -x;
    }
    return s;
  }
}

class Solution {
  public int bestSeqAtIndex(int[] height, int[] weight) {
    int n = height.length;
    int[][] arr = new int[n][2];
    for (int i = 0; i < n; ++i) {
      arr[i] = new int[] {height[i], weight[i]};
    }
    Arrays.sort(arr, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
    Set<Integer> s = new HashSet<>();
    for (int[] e : arr) {
      s.add(e[1]);
    }
    List<Integer> alls = new ArrayList<>(s);
    Collections.sort(alls);
    Map<Integer, Integer> m = new HashMap<>(alls.size());
    for (int i = 0; i < alls.size(); ++i) {
      m.put(alls.get(i), i + 1);
    }
    BinaryIndexedTree tree = new BinaryIndexedTree(alls.size());
    int ans = 1;
    for (int[] e : arr) {
      int x = m.get(e[1]);
      int t = tree.query(x - 1) + 1;
      ans = Math.max(ans, t);
      tree.update(x, t);
    }
    return ans;
  }
}

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

  void update(int x, int val) {
    while (x <= n) {
      c[x] = max(c[x], val);
      x += x & -x;
    }
  }

  int query(int x) {
    int s = 0;
    while (x > 0) {
      s = max(s, c[x]);
      x -= x & -x;
    }
    return s;
  }

private:
  int n;
  vector<int> c;
};

class Solution {
public:
  int bestSeqAtIndex(vector<int>& height, vector<int>& weight) {
    int n = height.size();
    vector<pair<int, int>> people;
    for (int i = 0; i < n; ++i) {
      people.emplace_back(height[i], weight[i]);
    }
    sort(people.begin(), people.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
      if (a.first == b.first) {
        return a.second > b.second;
      }
      return a.first < b.first;
    });
    vector<int> alls = weight;
    sort(alls.begin(), alls.end());
    alls.erase(unique(alls.begin(), alls.end()), alls.end());
    BinaryIndexedTree tree(alls.size());
    int ans = 1;
    for (auto& [_, w] : people) {
      int x = lower_bound(alls.begin(), alls.end(), w) - alls.begin() + 1;
      int t = tree.query(x - 1) + 1;
      ans = max(ans, t);
      tree.update(x, t);
    }
    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) update(x, val int) {
  for x <= this.n {
    if this.c[x] < val {
      this.c[x] = val
    }
    x += x & -x
  }
}

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

func bestSeqAtIndex(height []int, weight []int) int {
  n := len(height)
  people := make([][2]int, n)
  s := map[int]bool{}
  for i := range people {
    people[i] = [2]int{height[i], weight[i]}
    s[weight[i]] = true
  }
  sort.Slice(people, func(i, j int) bool {
    a, b := people[i], people[j]
    return a[0] < b[0] || a[0] == b[0] && a[1] > b[1]
  })
  alls := make([]int, 0, len(s))
  for k := range s {
    alls = append(alls, k)
  }
  sort.Ints(alls)
  tree := newBinaryIndexedTree(len(alls))
  ans := 1
  for _, p := range people {
    x := sort.SearchInts(alls, p[1]) + 1
    t := tree.query(x-1) + 1
    ans = max(ans, t)
    tree.update(x, t)
  }
  return ans
}