检查是否有任何间隔完全重叠

间隔表示为开始时间和结束时间的组合。给定一组间隔，我们需要编写一个程序来检查是否有任何间隔完全重叠。

例子：

Input:  arr[] = {{1, 3}, {1, 7}, {4, 8}, {2, 5}}
Output: true
The intervals {1, 3} completely overlaps in {1, 7}. 

Input:  arr[] = {{1, 3}, {7, 9}, {4, 6}, {10, 13}}
Output: false
No pair of intervals overlap. 

一个简单的解决方案是考虑每对间隔，并检查该对是否重叠。该解决方案的时间复杂度为O(n ² )。

更好的解决方案是使用Sorting 。以下是完整的算法。
1)按照开始时间的升序对所有间隔进行排序。此步骤需要O(n Logn)时间。
2)在排序数组中，如果一个间隔的结束时间不超过前一个间隔的结束，则存在完全重叠。此步骤需要O(n)时间。

以下是上述方法的实现：

// A C++ program to check if any two intervals
// completely overlap
#include 
using namespace std;
  
// An interval has start time and end time
struct Interval {
    int start;
    int end;
};
  
// Compares two intervals according to their staring 
// time. This is needed for sorting the intervals 
// using library function std::sort().
bool compareInterval(Interval i1, Interval i2)
{
    return (i1.start < i2.start) ? true : false;
}
  
// Function to check if any two intervals 
// completely overlap
bool isOverlap(Interval arr[], int n)
{
    // Sort intervals in increasing order of
    // start time
    sort(arr, arr + n - 1, compareInterval);
  
    // In the sorted array, if end time of an 
    // interval is not more than that of
    // end of previous interval, then there
    // is an overlap
    for (int i = 1; i < n; i++)
        if (arr[i].end <= arr[i - 1].end)
            return true;
  
    // If we reach here, then no overlap
    return false;
}
  
// Driver code
int main()
{
    // 1st example
    Interval arr1[] = { { 1, 3 }, { 1, 7 }, { 4, 8 }, 
                                          { 2, 5 } };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    if (isOverlap(arr1, n1))
        cout << "Yes\n";
    else
        cout << "No\n";
  
    // 2nd example
    Interval arr2[] = { { 1, 3 }, { 7, 9 }, { 4, 6 }, 
                                         { 10, 13 } };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    if (isOverlap(arr2, n2))
        cout << "Yes\n";
    else
        cout << "No\n";
  
    return 0;
}

输出：

Yes
No