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  • 【leetcode刷题笔记】Merge Intervals

    Given a collection of intervals, merge all overlapping intervals.

    For example,
    Given [1,3],[2,6],[8,10],[15,18],
    return [1,6],[8,10],[15,18].


     

    题解:首先对所有的区间按照start大小排序,然后遍历排序后的数组,用last记录前一个区间,如果遍历的当前区间可以和last合并,就把它合并到last里面;否则就把last放到answer list中,并且更新last。

    代码如下:

     1 /**
     2  * Definition for an interval.
     3  * public class Interval {
     4  *     int start;
     5  *     int end;
     6  *     Interval() { start = 0; end = 0; }
     7  *     Interval(int s, int e) { start = s; end = e; }
     8  * }
     9  */
    10 public class Solution {
    11     class IntervalComparator implements Comparator<Interval>{
    12         public int compare(Interval a,Interval b){
    13             return a.start - b.start;
    14         }
    15     }
    16     public List<Interval> merge(List<Interval> intervals) {
    17         if(intervals == null || intervals.size() <= 1)
    18             return intervals;
    19         Collections.sort(intervals,new IntervalComparator());
    20         
    21         ArrayList<Interval> answer = new ArrayList<Interval>();
    22         
    23         Interval last = intervals.get(0);
    24         for(int i = 1;i < intervals.size();i++){
    25             Interval now = intervals.get(i);
    26             if(last.end >= now.start){
    27                 last.end = Math.max(last.end, now.end);
    28             }
    29             else{
    30                 answer.add(last);
    31                 last = now;
    32             }
    33         }
    34         answer.add(last);
    35         return answer;
    36         
    37          
    38     }
    39 }
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  • 原文地址:https://www.cnblogs.com/sunshineatnoon/p/3863631.html
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