435. Non-overlapping-Intervals
difficulty: Medium
section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }
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.
Method One
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if(intervals.length < 2) {
return 0;
}
Arrays.sort(intervals, (a,b) -> a[0] - b[0]);
int count = -1;
int curEnd = intervals[0][1];
for(int[] interval : intervals) {
if(interval[0] < curEnd) {
count++;
curEnd = Math.min( interval[1], curEnd );
}else{
curEnd = interval[1];
}
}
return count;
}
}
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