LeetCode435. Non-overlapping Intervals

一、题目

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Example 3:

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don’t need to remove any of the intervals since they’re already non-overlapping.

Constraints:

1 <= intervals.length <= 105
intervals[i].length == 2
-5 * 104 <= starti < endi <= 5 * 104

二、题解

class Solution {
   
public:
    static bool cmp(vector<int>& a,vector<int>& b){
   
        return a[0] < b[0];
    }
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
   
        sort(intervals.begin(),intervals.end(),cmp);
        int res = 0;
        for(int i = 1;i < intervals.size();i++){
   
            if(intervals[i][0] < intervals[i-1][1]){
   
                res++;
                intervals[i][1] = min(intervals[i-1][1],intervals[i][1]);
            }
        }
        return res;
    }
};