[leetcode] 435. Non-overlapping Intervals

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:

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.

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& intervals) {
        int res=0;
        int n=intervals.size();
        sort(intervals.begin(),intervals.end(),[](Interval a,Interval b){
            return a.start

参考文献

[LeetCode] Non-overlapping Intervals 非重叠区间