本文主要是介绍LeetCode435. Non-overlapping Intervals，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
文章目录
- 一、题目
- 二、题解
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;}
};
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