Leetcode: Non-overlapping Intervals
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: 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.
Actually, the problem is the same as "Given a collection of intervals, find the maximum number of intervals that are non-overlapping." (the classic Greedy problem: Interval Scheduling). With the solution to that problem, guess how do we get the minimum number of intervals to remove? : )
Sorting Interval.end in ascending order is O(nlogn), then traverse intervals array to get the maximum number of non-overlapping intervals is O(n). Total is O(nlogn).
开始的时候想岔了,以为是要求同一时刻overlap的最多interval数,但仔细想一想就发现不对,应该是non-overlap的interval的最大数目
1. Best solution: sorted by interval end
case 1 add current interval as another non-overlapping interval, case 2 and case 3 all get rid of the current interval
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class Solution {
public int eraseOverlapIntervals(Interval[] intervals) {
if (intervals.length == 0) return 0;
int nonOverlap = 1;
int seq = 0;
Arrays.sort(intervals, new Comparator<Interval>() {
public int compare(Interval i1, Interval i2) {
return i1.end - i2.end;
}
});
for (int i=1; i<intervals.length; i++) {
if (intervals[i].start >= intervals[seq].end) {
seq = i;
nonOverlap++;
}
}
return intervals.length - nonOverlap;
}
}
Comparator can also be rewritten as
Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[1], i2[1]));
2. Alternatives(not the best): sort by interval start
case 1 add current interval as another non-overlapping interval, case 2 update the previous non-overlapping interval with the current one, and case 3 get rid of the current interval. So more cases need to be processed than sorted by interval end
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if (intervals.length < 1) return 0;
int seq = 0;
int nonOverlap = 1; Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0])); for (int i = 0; i < intervals.length; i ++) {
if (intervals[i][0] >= intervals[seq][1]) {
seq = i;
nonOverlap ++;
}
else if (intervals[i][1] <= intervals[seq][1]) {
seq = i;
}
} return intervals.length - nonOverlap;
}
}
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