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.
题目链接：https://leetcode.com/problems/non-overlapping-intervals/description/
题目分析：先用贪心求最大的不重合区间个数（按end排序，扫一遍），再一减即可 （击败70%）
/**
* 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; }
* }
*/
class Solution {
public int eraseOverlapIntervals(Interval[] intervals) {
int n = intervals.length;
if (n == 0) {
return 0;
}
Comparator <Interval> cmp = new Comparator<Interval>() {
public int compare(Interval i1, Interval i2) {
if (i1.end > i2.end) {
return 1;
} else if (i1.end < i2.end) {
return -1;
} else {
if (i1.start > i2.start) {
return 1;
} else if (i1.start < i2.start) {
return -1;
} else {
return 0;
}
}
}
};
Arrays.sort(intervals, cmp);
int maxSeg = 1, cur = intervals[0].end, i = 1;
while (i < n) {
while (i < n && intervals[i].start < cur) {
i++;
}
if (i < n) {
cur = intervals[i].end;
maxSeg++;
}
i++;
}
return n - maxSeg;
}
}