LeetCode : 435. Non-overlapping Intervals 非重叠子区间

落日映苍穹つ 2021-06-24 16:10 378阅读 0赞

**试题**  
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.  
**代码**  
这题可以转为求解不重叠的区间个数。那么对于重叠的两个区间，我们应该选哪一个？显然，然后end较小的后能够在后面容纳更多的区间。贪心思想。

/**
     * 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) {
            if(intervals==null || intervals.length==0) return 0;
            
            Arrays.sort(intervals, new Comparator<Interval>(){
                public int compare(Interval o1, Interval o2){
                    return o1.end - o2.end;
                }
            });
            
            int count = 1;
            int minEnd = intervals[0].end;
            for(int i=1; i<intervals.length; i++){
                if(intervals[i].start<minEnd) continue;
                count++;
                minEnd = intervals[i].end;
            }
            return intervals.length-count;
        }
    }