leetcode 56 Merge Intervals
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 2018-05-20 11:04:18  

Given a collection of intervals, merge all overlapping intervals.

Example 1:

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Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

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Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considerred overlapping.
  • 先按照start从小到大的顺序排序
  • 设定当前start，end为第一个的值
  • 遍历一遍intervals，在这个过程中更新start，end，每次更新，就把原来的值插入到ans中
  • 注意每次比较的时候不是i不是和i-1去比较，而是和start，end比较
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/**
 * 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 List<Interval> merge(List<Interval> intervals) {
        ArrayList<Interval> ans = new ArrayList<Interval>();
        
        if(intervals.size()==0 || intervals == null){
            return ans;
        }
        Collections.sort(intervals, new Comparator<Interval>(){
            @Override
            public int compare(Interval o1, Interval o2){
                    if(o1.start < o2.start){
                    return -1;
                }
                else if(o1.start > o2.start){
                    return 1;
                }
                else{
                    return 0;
                }
            }
        });
        int start = intervals.get(0).start;
        int end = intervals.get(0).end;
        for(int i=1; i< intervals.size(); i++){
            if(intervals.get(i).start <= end && intervals.get(i).end <= end){
                         
            }
            else if(intervals.get(i).start <= end && intervals.get(i).end > end){
                end = intervals.get(i).end;
            }
            else{
                ans.add(new Interval(start, end));
                start = intervals.get(i).start;
                end = intervals.get(i).end;
            }
            
        }
        ans.add(new Interval(start, end));
        return ans;
        
    }
}

讨论区更descent的方法：

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class Solution {
    private class IntervalComparator implements Comparator<Interval> {
        @Override
        public int compare(Interval a, Interval b) {
            return a.start < b.start ? -1 : a.start == b.start ? 0 : 1;
        }
    }

    public List<Interval> merge(List<Interval> intervals) {
        Collections.sort(intervals, new IntervalComparator());

        LinkedList<Interval> merged = new LinkedList<Interval>();
        for (Interval interval : intervals) {
            // if the list of merged intervals is empty or if the current
            // interval does not overlap with the previous, simply append it.
            if (merged.isEmpty() || merged.getLast().end < interval.start) {
                merged.add(interval);
            }
            // otherwise, there is overlap, so we merge the current and previous
            // intervals.
            else {
                merged.getLast().end = Math.max(merged.getLast().end, interval.end);
            }
        }

        return merged;
    }