LeetCode #435 - Non-overlapping Intervals

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:

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.

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& 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