435. Non-overlapping Intervals

435. Non-overlapping Intervals

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

Example 2:

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

Example 3:

Input: [ [1,2], [2,3] ]Output: 0Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

思路：贪心算法，以每个元素的end为关键元素排序，end相等的情况下，按start升序排序。这里的实际类型和这个问题很像，

只有一个场地，要安排尽量多的活动。贪心规则：尽量安排结束时间早的活动。如果后面的活动与已经安排好的兼容，则加入集合。

/** * Definition for an interval. * struct Interval { *     int start; *     int end; *     Interval() : start(0), end(0) {} *     Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {    public:    int eraseOverlapIntervals(vector<Interval>& intervals) {       int n=intervals.size();       if(!n) return 0;       sort(intervals.begin(),intervals.end(),[](Interval a,Interval b){           return a.end<b.end||(a.start>b.start&&a.end==b.end);       });       int cnt=1, s=intervals[0].start,e=intervals[0].end;       for(int i=1;i<n;i++)       {           //if(it.start==s&&it.end>e) continue;           Interval it=intervals[i];           if(it.start>=e) {               cnt++;              // s=it.start;               e=it.end;           }       }       return n-cnt;    }};