合并重叠区间

给定一组任意顺序的时间间隔，将所有重叠的时间间隔合并为一个，并输出应该只有互斥时间间隔的结果。为简单起见，将区间表示为整数对。
例如，让给定的区间集是 {{1,3}, {2,4}, {5,7}, {6,8}}。区间 {1,3} 和 {2,4} 相互重叠，因此它们应该合并成为 {1, 4}。同样，{5, 7} 和 {6, 8} 应该合并成为 {5, 8}

编写一个函数，为给定的间隔集生成合并的间隔集。
一个简单的方法是从第一个区间开始，将其与所有其他区间进行重叠比较，如果它与任何其他区间重叠，则从列表中删除另一个区间并将另一个区间合并到第一个区间中。在第一次之后对剩余的间隔重复相同的步骤。这种方法不能在 O(n^2) 时间内更好地实施。
一种有效的方法是首先根据开始时间对区间进行排序。一旦我们有了排序的区间，我们就可以将所有区间组合成一个线性遍历。这个想法是，在已排序的区间数组中，如果区间 [i] 不与区间 [i-1] 重叠，则区间 [i+1] 不能与区间 [i-1] 重叠，因为区间 [i 的开始时间+1] 必须大于或等于区间[i]。以下是详细的逐步算法。

1. Sort the intervals based on increasing order of 
    starting time.
2. Push the first interval on to a stack.
3. For each interval do the following
   a. If the current interval does not overlap with the stack 
       top, push it.
   b. If the current interval overlaps with stack top and ending
       time of current interval is more than that of stack top, 
       update stack top with the ending  time of current interval.
4. At the end stack contains the merged intervals. 

下面是上述方法的一个实现。

C++

// A C++ program for merging overlapping intervals
#include
using namespace std;
 
// An interval has start time and end time
struct Interval
{
    int start, end;
};
 
// Compares two intervals according to their starting time.
// This is needed for sorting the intervals using library
// function std::sort(). See http://goo.gl/iGspV
bool compareInterval(Interval i1, Interval i2)
{
    return (i1.start < i2.start);
}
 
// The main function that takes a set of intervals, merges
// overlapping intervals and prints the result
void mergeIntervals(Interval arr[], int n)
{
    // Test if the given set has at least one interval
    if (n <= 0)
        return;
 
    // Create an empty stack of intervals
    stack s;
 
    // sort the intervals in increasing order of start time
    sort(arr, arr+n, compareInterval);
 
    // push the first interval to stack
    s.push(arr[0]);
 
    // Start from the next interval and merge if necessary
    for (int i = 1 ; i < n; i++)
    {
        // get interval from stack top
        Interval top = s.top();
 
        // if current interval is not overlapping with stack top,
        // push it to the stack
        if (top.end < arr[i].start)
            s.push(arr[i]);
 
        // Otherwise update the ending time of top if ending of current
        // interval is more
        else if (top.end < arr[i].end)
        {
            top.end = arr[i].end;
            s.pop();
            s.push(top);
        }
    }
 
    // Print contents of stack
    cout << "\n The Merged Intervals are: ";
    while (!s.empty())
    {
        Interval t = s.top();
        cout << "[" << t.start << "," << t.end << "] ";
        s.pop();
    }
    return;
}
 
// Driver program
int main()
{
    Interval arr[] =  { {6,8}, {1,9}, {2,4}, {4,7} };
    int n = sizeof(arr)/sizeof(arr[0]);
    mergeIntervals(arr, n);
    return 0;
}

Java

// A Java program for merging overlapping intervals
import java.util.Arrays;
import java.util.Comparator;
import java.util.Stack;
public class MergeOverlappingIntervals {
 
    // The main function that takes a set of intervals, merges
    // overlapping intervals and prints the result
    public static void mergeIntervals(Interval arr[])
    {
        // Test if the given set has at least one interval
        if (arr.length <= 0)
            return;
   
        // Create an empty stack of intervals
        Stack stack=new Stack<>();
   
        // sort the intervals in increasing order of start time
        Arrays.sort(arr,new Comparator(){
            public int compare(Interval i1,Interval i2)
            {
                return i1.start-i2.start;
            }
        });
   
        // push the first interval to stack
        stack.push(arr[0]);
   
        // Start from the next interval and merge if necessary
        for (int i = 1 ; i < arr.length; i++)
        {
            // get interval from stack top
            Interval top = stack.peek();
   
            // if current interval is not overlapping with stack top,
            // push it to the stack
            if (top.end < arr[i].start)
                stack.push(arr[i]);
   
            // Otherwise update the ending time of top if ending of current
            // interval is more
            else if (top.end < arr[i].end)
            {
                top.end = arr[i].end;
                stack.pop();
                stack.push(top);
            }
        }
   
        // Print contents of stack
        System.out.print("The Merged Intervals are: ");
        while (!stack.isEmpty())
        {
            Interval t = stack.pop();
            System.out.print("["+t.start+","+t.end+"] ");
        } 
    } 
 
    public static void main(String args[]) {
        Interval arr[]=new Interval[4];
        arr[0]=new Interval(6,8);
        arr[1]=new Interval(1,9);
        arr[2]=new Interval(2,4);
        arr[3]=new Interval(4,7);
        mergeIntervals(arr);
    }
}
 
class Interval
{
    int start,end;
    Interval(int start, int end)
    {
        this.start=start;
        this.end=end;
    }
}
// This code is contributed by Gaurav Tiwari

C#

// A C# program for merging overlapping intervals
using System;
using System.Collections;
using System.Collections.Generic;
 
public class MergeOverlappingIntervals
{
 
  // sort the intervals in increasing order of start time
  class sortHelper : IComparer
  {
    int IComparer.Compare(object a, object b)
    {
      Interval first = (Interval)a;
      Interval second = (Interval)b;
      if (first.start == second.start)
      {
        return first.end - second.end;
      }
      return first.start - second.start;
    }
  }
 
  // The main function that takes a set of intervals, merges
  // overlapping intervals and prints the result
  public static void mergeIntervals(Interval []arr)
  {
 
    // Test if the given set has at least one interval
    if (arr.Length <= 0)
      return;
    Array.Sort(arr, new sortHelper());
 
    // Create an empty stack of intervals
    Stack stack = new Stack();
 
    // Push the first interval to stack
    stack.Push(arr[0]);
 
    // Start from the next interval and merge if necessary
    for (int i = 1 ; i < arr.Length; i++)
    {
 
      // get interval from stack top
      Interval top = (Interval)stack.Peek();
 
      // if current interval is not overlapping with stack top,
      // Push it to the stack
      if (top.end < arr[i].start)
        stack.Push(arr[i]);
 
      // Otherwise update the ending time of top if ending of current
      // interval is more
      else if (top.end < arr[i].end)
      {
        top.end = arr[i].end;
        stack.Pop();
        stack.Push(top);
      }
    }
 
    // Print contents of stack
    Console.Write("The Merged Intervals are: ");
    while (stack.Count != 0)
    {
      Interval t = (Interval)stack.Pop();
      Console.Write("[" + t.start + "," + t.end + "] ");
    }
  }
 
  // Driver code
  public static void Main()
  {
 
    Interval []arr = new Interval[4];
    arr[0] = new Interval(6, 8);
    arr[1] = new Interval(1, 9);
    arr[2] = new Interval(2, 4);
    arr[3] = new Interval(4, 7);
    mergeIntervals(arr);
  }
}
 
public class Interval
{
  public int start,end;
  public Interval(int start, int end)
  {
    this.start = start;
    this.end = end;
  }
}
 
// This code is contributed by rutvik_56.

输出：

The Merged Intervals are: [1,9] 

该方法的时间复杂度为 O(nLogn)，用于排序。对区间数组进行排序后，合并需要线性时间。
AO(n Log n) 和 O(1) 额外空间解决方案
上述解决方案需要 O(n) 额外的堆栈空间。我们可以通过就地进行合并操作来避免使用额外的空间。下面是详细步骤。

1) Sort all intervals in increasing order of start time.
2) Traverse sorted intervals starting from first interval, 
   do following for every interval.
      a) If current interval is not first interval and it 
         overlaps with previous interval, then merge it with
         previous interval. Keep doing it while the interval
         overlaps with the previous one.         
      b) Else add current interval to output list of intervals.

请注意，如果区间按开始时间的降序排序，我们可以通过比较前一个区间的开始时间和当前区间的结束时间来快速检查区间是否重叠。
下面是上述算法的实现。

C++

// C++ program to merge overlapping Intervals in
// O(n Log n) time and O(1) extra space.
#include
using namespace std;
 
// An Interval
struct Interval
{
    int s, e;
};
 
// Function used in sort
bool mycomp(Interval a, Interval b)
{ return a.s < b.s; }
 
void mergeIntervals(Interval arr[], int n)
{
    // Sort Intervals in increasing order of
    // start time
    sort(arr, arr+n, mycomp);
 
    int index = 0; // Stores index of last element
    // in output array (modified arr[])
 
    // Traverse all input Intervals
    for (int i=1; i=  arr[i].s)
        {
               // Merge previous and current Intervals
            arr[index].e = max(arr[index].e, arr[i].e);
        }
        else {
            index++;
            arr[index] = arr[i];
        }   
    }
 
    // Now arr[0..index-1] stores the merged Intervals
    cout << "\n The Merged Intervals are: ";
    for (int i = 0; i <= index; i++)
        cout << "[" << arr[i].s << ", " << arr[i].e << "] ";
}
 
// Driver program
int main()
{
    Interval arr[] = { {6,8}, {1,9}, {2,4}, {4,7} };
    int n = sizeof(arr)/sizeof(arr[0]);
    mergeIntervals(arr, n);
    return 0;
}

Java

// Java program to merge overlapping Intervals in
// O(n Log n) time and O(1) extra space
 
import java.util.Arrays;
import java.util.Comparator;
 
// An Interval
class Interval
{
    int start,end;
     
    Interval(int start, int end)
    {
        this.start=start;
        this.end=end;
    }
}
 
public class MergeOverlappingIntervals {
     
    // Function that takes a set of intervals, merges
    // overlapping intervals and prints the result
    public static void mergeIntervals(Interval arr[])
    {
        // Sort Intervals in increasing order of
        // start time
        Arrays.sort(arr,new Comparator(){
            public int compare(Interval i1,Interval i2)
            {
                return i1.start - i2.start;
            }
        });
   
        int index = 0; // Stores index of last element
        // in output array (modified arr[])
   
        // Traverse all input Intervals
        for (int i=1; i=  arr[i].start)
            {
                   // Merge previous and current Intervals
                arr[index].end = Math.max(arr[index].end, arr[i].end);
            }
            else {
                index++;
                arr[index] = arr[i];
            }   
        }
         
        // Now arr[0..index-1] stores the merged Intervals
        System.out.print("The Merged Intervals are: ");
        for (int i = 0; i <= index; i++)
        {
            System.out.print("[" + arr[i].start + ","
                                        + arr[i].end + "]");
        }
    }
 
    // Driver Code
    public static void main(String args[]) {
        Interval arr[]=new Interval[4];
        arr[0]=new Interval(6,8);
        arr[1]=new Interval(1,9);
        arr[2]=new Interval(2,4);
        arr[3]=new Interval(4,7);
        mergeIntervals(arr);
    }
}
 
// This code is contributed by Gaurav Tiwari
// This code was fixed by Subham Mukhopadhyay

Python3

# Python3 program to merge overlapping Intervals
# in O(n Log n) time and O(1) extra space
def mergeIntervals(arr):
         
        # Sorting based on the increasing order
        # of the start intervals
        arr.sort(key = lambda x: x[0])
         
        # array to hold the merged intervals
        m = []
        s = -10000
        max = -100000
        for i in range(len(arr)):
            a = arr[i]
            if a[0] > max:
                if i != 0:
                    m.append([s,max])
                max = a[1]
                s = a[0]
            else:
                if a[1] >= max:
                    max = a[1]
         
        #'max' value gives the last point of
        # that particular interval
        # 's' gives the starting point of that interval
        # 'm' array contains the list of all merged intervals
 
        if max != -100000 and [s, max] not in m:
            m.append([s, max])
        print("The Merged Intervals are :", end = " ")
        for i in range(len(m)):
            print(m[i], end = " ")
 
# Driver code
arr = [[6, 8], [1, 9], [2, 4], [4, 7]]
mergeIntervals(arr)
 
# This code is contributed
# by thirumalai srinivasan

输出：

The Merged Intervals are: [1,9] 

感谢 Gaurav Ahirwar 提出这种方法。