给定N个范围[L，R]和整数K ，任务是检查是否有K个范围在任何点重叠。
例子：
Input: ranges[][] = {{1, 3}, {2, 4}, {3, 4}, {7, 10}}, K = 3
Output: Yes
3 is a common point among the
ranges {1, 3}, {2, 4} and {3, 4}.
Input: ranges[][] = {{1, 2}, {3, 4}, {5, 6}, {7, 8}}, K = 2
Output: No
方法：想法是制作一个向量对，并将每个范围的起点成对存储在该向量中，将其作为(起点-1)，将终点作为(终点1)。现在，对向量进行排序，然后遍历向量，如果当前元素是起点，则将其推入堆栈中；如果当前元素是终点，则从堆栈中弹出一个元素。如果在任何时间实例中，堆栈的大小都大于或等于K，则打印“是”，否则最后打印“否”。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Comparator to sort the vector of pairs
bool sortby(const pair& a,
const pair& b)
{
if (a.first != b.first)
return a.first < b.first;
return (a.second < b.second);
}
// Function that returns true if any k
// segments overlap at any point
bool kOverlap(vector > pairs, int k)
{
// Vector to store the starting point
// and the ending point
vector > vec;
for (int i = 0; i < pairs.size(); i++) {
// Starting points are marked by -1
// and ending points by +1
vec.push_back({ pairs[i].first, -1 });
vec.push_back({ pairs[i].second, +1 });
}
// Sort the vector by first element
sort(vec.begin(), vec.end());
// Stack to store the overlaps
stack > st;
for (int i = 0; i < vec.size(); i++) {
// Get the current element
pair cur = vec[i];
// If it is the starting point
if (cur.second == -1) {
// Push it in the stack
st.push(cur);
}
// It is the ending point
else {
// Pop an element from stack
st.pop();
}
// If more than k ranges overlap
if (st.size() >= k) {
return true;
}
}
return false;
}
// Driver code
int main()
{
vector > pairs;
pairs.push_back(make_pair(1, 3));
pairs.push_back(make_pair(2, 4));
pairs.push_back(make_pair(3, 5));
pairs.push_back(make_pair(7, 10));
int n = pairs.size(), k = 3;
if (kOverlap(pairs, k))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Stack;
class GFG{
static class Pair
{
int first, second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function that returns true if any k
// segments overlap at any point
static boolean kOverlap(ArrayList pairs,
int k)
{
// Vector to store the starting point
// and the ending point
ArrayList vec = new ArrayList<>();
for(int i = 0; i < pairs.size(); i++)
{
// Starting points are marked by -1
// and ending points by +1
vec.add(new Pair(pairs.get(i).first, -1));
vec.add(new Pair(pairs.get(i).second, +1));
}
// Sort the vector by first element
Collections.sort(vec, new Comparator()
{
// Comparator to sort the vector of pairs
public int compare(Pair a, Pair b)
{
if (a.first != b.first)
return a.first - b.first;
return (a.second - b.second);
}
});
// Stack to store the overlaps
Stack st = new Stack<>();
for(int i = 0; i < vec.size(); i++)
{
// Get the current element
Pair cur = vec.get(i);
// If it is the starting point
if (cur.second == -1)
{
// Push it in the stack
st.push(cur);
}
// It is the ending point
else
{
// Pop an element from stack
st.pop();
}
// If more than k ranges overlap
if (st.size() >= k)
{
return true;
}
}
return false;
}
// Driver code
public static void main(String[] args)
{
ArrayList pairs = new ArrayList<>();
pairs.add(new Pair(1, 3));
pairs.add(new Pair(2, 4));
pairs.add(new Pair(3, 5));
pairs.add(new Pair(7, 10));
int n = pairs.size(), k = 3;
if (kOverlap(pairs, k))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by sanjeev2552 Python3
# Python3 implementation of the approach
# Function that returns true if any k
# segments overlap at any point
def kOverlap(pairs: list, k):
# Vector to store the starting point
# and the ending point
vec = list()
for i in range(len(pairs)):
# Starting points are marked by -1
# and ending points by +1
vec.append((pairs[0], -1))
vec.append((pairs[1], 1))
# Sort the vector by first element
vec.sort(key = lambda a: a[0])
# Stack to store the overlaps
st = list()
for i in range(len(vec)):
# Get the current element
cur = vec[i]
# If it is the starting point
if cur[1] == -1:
# Push it in the stack
st.append(cur)
# It is the ending point
else:
# Pop an element from stack
st.pop()
# If more than k ranges overlap
if len(st) >= k:
return True
return False
# Driver Code
if __name__ == "__main__":
pairs = list()
pairs.append((1, 3))
pairs.append((2, 4))
pairs.append((3, 5))
pairs.append((7, 10))
n = len(pairs)
k = 3
if kOverlap(pairs, k):
print("Yes")
else:
print("No")
# This code is contributed by
# sanjeev2552C#
// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Function that returns true if any k
// segments overlap at any point
static bool kOverlap(List> pairs,
int k)
{
// Vector to store the starting point
// and the ending point
List> vec = new List>();
for(int i = 0; i < pairs.Count; i++)
{
// Starting points are marked by -1
// and ending points by +1
vec.Add(new Tuple(pairs[i].Item1,-1));
vec.Add(new Tuple(pairs[i].Item2,1));
}
vec.Sort();
// Stack to store the overlaps
Stack st = new Stack();
for(int i = 0; i < vec.Count; i++)
{
// Get the current element
Tuple cur = vec[i];
// If it is the starting point
if (cur.Item2 == -1)
{
// Push it in the stack
st.Push(cur);
}
// It is the ending point
else
{
// Pop an element from stack
st.Pop();
}
// If more than k ranges overlap
if (st.Count >= k)
{
return true;
}
}
return false;
}
// Driver code
public static void Main(params string[] args)
{
List> pairs = new List>();
pairs.Add(new Tuple(1, 3));
pairs.Add(new Tuple(2, 4));
pairs.Add(new Tuple(3, 5));
pairs.Add(new Tuple(7, 10));
int n = pairs.Count, k = 3;
if (kOverlap(pairs, k))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by rutvik_56/ 输出：
Yes