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
我们先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。
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32 | 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
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55 | 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;
}
}
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55 | 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;
}
};
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57 | 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
}
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59 | class BinaryIndexedTree {
private var n: Int
private var c: [Int]
init(_ n: Int) {
self.n = n
self.c = [Int](repeating: 0, count: n + 1)
}
func update(_ x: Int, _ val: Int) {
var x = x
while x <= n {
c[x] = max(c[x], val)
x += x & -x
}
}
func query(_ x: Int) -> Int {
var x = x
var s = 0
while x > 0 {
s = max(s, c[x])
x -= x & -x
}
return s
}
}
class Solution {
func bestSeqAtIndex(_ height: [Int], _ weight: [Int]) -> Int {
let n = height.count
var arr: [(Int, Int)] = []
for i in 0..<n {
arr.append((height[i], weight[i]))
}
arr.sort {
if $0.0 == $1.0 {
return $1.1 < $0.1
}
return $0.0 < $1.0
}
let weights = Set(arr.map { $1 })
let sortedWeights = Array(weights).sorted()
let m = sortedWeights.enumerated().reduce(into: [Int: Int]()) {
$0[$1.element] = $1.offset + 1
}
let tree = BinaryIndexedTree(sortedWeights.count)
var ans = 1
for (_, w) in arr {
let x = m[w]!
let t = tree.query(x - 1) + 1
ans = max(ans, t)
tree.update(x, t)
}
return ans
}
}
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