LeetCode 435. Non-overlapping Intervals
原题链接在这里:https://leetcode.com/problems/non-overlapping-intervals/
题目:
Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
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:
- 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.
题解:
Draw with some simple examples and find routine.
Sort based on ending point. Have max initialized as intervals[0] end.
Starting from 2nd, if current interval start is smaller than max, there is overlap, res++.
Otherwise, there is no overlap, we update the maximum.
Time Complexity: O(nlogn). n = intervals.length.
Space: O(1).
AC Java:
1 class Solution { 2 public int eraseOverlapIntervals(int[][] intervals) { 3 if(intervals == null || intervals.length < 2){ 4 return 0; 5 } 6 7 Arrays.sort(intervals, (a, b) -> a[1] == b[1] ? b[0] - a[0] : a[1] - b[1]); 8 int max = intervals[0][1]; 9 int res = 0; 10 11 for(int i = 1; i<intervals.length; i++){ 12 if(intervals[i][0] < max){ 13 res++; 14 }else{ 15 max = intervals[i][1]; 16 } 17 } 18 19 return res; 20 } 21 }