435. Non-overlapping Intervals

题目：

435. Non-overlapping Intervals

• Total Accepted: 10154 
• Total Submissions: 25196 
• Difficulty: Medium
• Contributors: love_Fawn

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.

解题思路：

此题和上次做的Minimum Number of Arrows to Burst Balloons是类似的，这题和上课讲的开会的例子就基本一样， 本题要求删除最少使得剩下的不重叠，开会的例子是要求最多的不重叠的数目，其实本质是一样的。做法都是先按照终止点从小到大排序，再比较每一个的起点是否小于上一个终止点，如果是则与上一个重叠，需要删除；如果不是，则不需要删除，上一个终止点更改为这个的终止点。代码也和上次做的题目差不多，再做一次类似的题目就当熟悉一下吧...

代码：

/** * 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) {    if(intervals.size() == 0)    return 0;    int count = 0;        sort(intervals.begin(), intervals.end(), comp);        int last_end = intervals[0].end;        for(int i = 1; i < intervals.size(); i++){        if(intervals[i].start < last_end){        count++;        continue;        }        last_end = intervals[i].end;        }        return count;    }    static bool comp(Interval a, Interval b){    return a.end == b.end ? a.start < b.start : a.end < b.end;    }};