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Description
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.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
分析
题目的意思是：给你很多个区间，求出移除区间后使得剩下的区间不重叠，求移除的最少区间的数量。
-先按照start排序
- 判断方法是看如果前一个区间的end大于后一个区间的start，那么一定是重复区间，此时我们结果res自增1，我们需要删除一个，那么此时我们究竟该删哪一个呢，为了保证我们总体去掉的区间数最小，我们去掉那个end值较大的区间，而在代码中，我们并没有真正的删掉某一个区间，而是用一个变量last指向上一个需要比较的区间，我们将last指向end值较小的那个区间；如果两个区间没有重叠，那么此时last指向当前区间，继续进行下一次遍历。
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) {
int res=0;
int n=intervals.size();
sort(intervals.begin(),intervals.end(),[](Interval a,Interval b){
return a.start<b.start;
});
int last=0;
for(int i=1;i<n;i++){
if(intervals[i].start<intervals[last].end){
res++;
if(intervals[i].end<intervals[last].end) last=i;
}else{
last=i;
}
}
return res;
}
};