435. Non-overlapping Intervals _intervals

行是知之始,知是行之成。这篇文章主要讲述435. Non-overlapping Intervals相关的知识，希望能为你提供帮助。
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: 1Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

【435. Non-overlapping Intervals】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.

求移除多少区间后，剩余区间都是不重叠的。

先求最多能组成多少不重叠的区间，再用总区间数减去不重叠区间数。

C++：

1 /** 2* Definition for an interval. 3* struct Interval { 4*int start; 5*int end; 6*Interval() : start(0), end(0) {} 7*Interval(int s, int e) : start(s), end(e) {} 8* }; 9*/ 10 bool compare(const Interval& a ,const Interval& b){ 11return a.end < b.end ; 12 } 13 class Solution { 14 public: 15int eraseOverlapIntervals(vector< Interval> & intervals) { 16if (intervals.size() == 0){ 17return 0 ; 18} 19sort(intervals.begin() , intervals.end() , compare) ; 20int cnt = 1; 21int end = intervals[0].end ; 22for(int i = 1 ; i < intervals.size() ; i++){ 23if (intervals[i].start < end){ 24continue ; 25} 26end = intervals[i].end ; 27cnt++ ; 28} 29return intervals.size() - cnt ; 30} 31 };