• leetcode 56 Merge Intervals ----- java


    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].

    这道题就是合并所有集合,虽然是hard难度,但是还是比较好想出来的,就是重写compare,然后对所有集合进行排序,排序之后再进行比较久ok了。

    [A,B]与[C,D]

    在已经排序的前提下,比较B与C的大小,如果B大,那么合并两个集合,取出D与下一个集合进行比较

                      如果C大,那么直接添加集合进result。

    最后再进行一次判断(最后一个集合是否加入了result中)。

    但是第一次提交compare定义出现了错误。

    /**
     * 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 List<Interval> merge(List<Interval> intervals) {
            List<Interval> result = new ArrayList<Interval>();
            int len = intervals.size();
            if( len < 2 )
                return intervals;
            Comparator<Interval> comparator = new Comparator<Interval>(){
                public int compare(Interval i1,Interval i2){
                    if( i1.start>i2.start)
                        return 1;
                    else 
                        return -1;
                }
            };
            Collections.sort(intervals, comparator);
            int[] start = new int[len];
            int[] end = new int[len];
            for( int i = 0;i<len;i++){
                start[i] = intervals.get(i).start;
                end[i] = intervals.get(i).end;
            }
            int begin,over;
            int i = 0,j = 1;
            while( i< len ){
                if( i == len-1){
                    result.add(intervals.get(i));
                    return result;
                }
                Interval ans = new Interval();
                ans.start = start[i];
                begin = start[j];
                over = end[i];
                while( over >= begin ){
                    over = over>end[j]?over:end[j];
                    j++;
                    if( j == len)
                        break;
                    begin = start[j];
                }
                ans.end = over;
                result.add(ans);
                
                i = j;
                j = j+1;
            }
            
            
            return result;
        }
    }

    然后发现并不是超时,只是对于compare重写的时候出现了问题。

    jdk  1.7以前compare只能返回1与-1,但是1.7以后当两个数相同需要返回0,因为compare(a,b)与compare(b,a)要返回相反的数,所以这里出现了问题,然后稍微修改一下,并且做了一些细微的调整,结果还算满意。

    /**
     * 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 List<Interval> merge(List<Interval> intervals) {
            List<Interval> result = new ArrayList<Interval>();
            int len = intervals.size();
            if( len < 2 )
                return intervals;
            Comparator<Interval> comparator = new Comparator<Interval>(){
                public int compare(Interval i1,Interval i2){
                    if( i1.start>i2.start)
                        return 1;
                    else if( i1.start == i2.start)
                        return 0;
                    else
                        return -1;
                }
            };
            Collections.sort(intervals, comparator);
            int begin,over;
            int i = 0,j = 1,k = 0;
            while( i< len && j < len){
                Interval ans = new Interval();
                ans.start = intervals.get(i).start;
                begin = intervals.get(j).start;
                over = intervals.get(i).end;
                while( over >= begin ){
                    over = over>intervals.get(j).end?over:intervals.get(j).end;
                    j++;
                    if( j == len)
                        break;
                    begin = intervals.get(j).start;
                }
                ans.end = over;
                result.add(ans);
                i = j;
                j = j+1;
                k++;
            }
            if( result.get(k-1).end < intervals.get(len-1).start )
                result.add(intervals.get(len-1));
            
            return result;
        }
    }

     然后发现别人的答案中,有一个很难发现的细节,就是如果将start和end分别放入两个数组中,再进行排序,最后的结果与直接将intervals排序得到的结果一样,而这样做的话,就直接达到了最快。其实想法是一样的,还是以前说的,算法是一方面,另一个方面就是存储方式。两者都很关键。

    /**
     * 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 List<Interval> merge(List<Interval> intervals) {
            List<Interval> result = new ArrayList<Interval>();
            int len = intervals.size();
            if( len < 2 )
                return intervals;
            int[] start = new int[len];
            int[] end = new int[len];
            for( int i = 0;i<len;i++){
                start[i] = intervals.get(i).start;
                end[i] = intervals.get(i).end;
            }
            Arrays.sort(start);
            Arrays.sort(end);
            int begin,over;
            int i = 0,j = 1;
            while( i< len ){
                if( i == len-1){
                    result.add(new Interval(start[i],end[i]));
                    return result;
                }
                Interval ans = new Interval();
                ans.start = start[i];
                begin = start[j];
                over = end[i];
                while( over >= begin ){
                    over = over>end[j]?over:end[j];
                    j++;
                    if( j == len)
                        break;
                    begin = start[j];
                }
                ans.end = over;
                result.add(ans);
                
                i = j;
                j = j+1;
            }
            
            
            return result;
        }
    }
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  • 原文地址:https://www.cnblogs.com/xiaoba1203/p/5761729.html
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