Leetcode 435. Non-overlapping Intervals题解

435. Non-overlapping Intervals

 

Question Editorial Solution

  My Submissions

• Total Accepted: 1112
• Total Submissions: 3007
• Difficulty: Medium
• Contributors: love_FDU_llp

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Example 1:

Input: [ [1,2], [2,3], [3,4], [1,3] ]

Output: 1

Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:

Input: [ [1,2], [1,2], [1,2] ]

Output: 2

Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

解题思路：

本题即为在一个时间段内能安排最大活动数的简单变形。应用贪心的思想，首先将所有的interval按照结束时间排序，然后找到所有时间不重叠的interval。再用interval的总数减去不重叠的interval的数目即可。

代码展示：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
 #include 
 using namespace std;
class Solution {
public:
    bool cmp(Interval t1 , Interval t2)
    {
        return t1.end& intervals,int n)
    {
        for(int i=0;i& intervals) {
        int n= intervals.size();
        if(!n) return 0;
        mysort(intervals,n);
        int count=1;
        int e = intervals[0].end;
        for(int i=1;i=e)
            {
                count++;
                e=intervals[i].end;
            }
        }
        return n-count;
    }
};