LeetCode #435 - Non-overlapping Intervals

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本文介绍了一种算法，用于解决给定一系列区间时如何最小化地移除某些区间以确保所有区间都不相交的问题。该算法首先按区间起始点排序，接着通过遍历并维护当前区间的最大终点来判断是否需要移除区间。

题目描述：

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:

You may assume the interval's end point is always bigger than its start point.
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.

给出一组区间，求问需要最少移除多少个区间，才能让它们不相交。首先需要对区间按照起点排序，然后依次遍历各个区间，同时维护一个当前所有区间的终点，这个终点必然是所有区间终点的最大值。当遍历到一个新的区间之后，如果它的起点小于当前终点，说明产生了相交，需要移除一个区间，自然就是移除那个终点较大的区间，同时更新当前终点。

class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        if(intervals.size()<=1) return 0;
        sort(intervals.begin(),intervals.end(),comp);
        int count=0;
        int cur_end=intervals[0].end;
        for(int i=1;i<intervals.size();i++)
        {
            if(intervals[i].start<cur_end)
            {
                count++;
                cur_end=min(cur_end,intervals[i].end);
            }
            else cur_end=intervals[i].end;
        }
        return count;
    }
    
    static bool comp(Interval a, Interval b)
    {
        if(a.start<=b.start) return true;
        else return false;
    }
};