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  • 435. 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:

    1. You may assume the interval's end point is always bigger than its start point.
    2. 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.

    /**
     * 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;
            Arrays.sort(intervals, new Comparator<Interval>(){
                public int compare(Interval a, Interval b){
                    return a.start - b.start;
                }
            });
            int res = 0;
            int min = intervals[0].end;
            for(int i = 1 ; i < intervals.length ; i++){
                if(intervals[i].start < min){
                    res++;
                    min = Math.min(min, intervals[i].end);  // key point to make sure minimum : del the inteval with large end
                }
                else
                  min = intervals[i].end;
            }
            return res;
        }
    }
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  • 原文地址:https://www.cnblogs.com/joannacode/p/6108943.html
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