Hard-题目28：57. Insert Interval

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本文介绍了一个关于区间集合的问题，即如何插入一个新的区间并进行合并化简。通过对比Merge Intervals问题，提供了一种解决方案，包括定义区间比较器，并实现区间插入和合并的功能。

题目原文：
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
题目大意：
给出一个区间集合，插入一个区间并化简。
题目分析：
跟前面的Hard-题目22：56. Merge Intervals是一回事，但是数据强了一些，要把数组开到40000.
源码：（language：java）

/**
 * 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 {
    private class IntervalComparator implements Comparator<Interval> {
        @Override
        public int compare(Interval o1, Interval o2) {
            // TODO Auto-generated method stub
            if(o1.start<o2.start)
                return -1;
            else if(o1.start>o2.start)
                return 1;
            else {
                if(o1.end<o2.end)
                    return -1;
                else if(o1.end>o2.end)
                    return 1;
                else {
                    return 0;
                }
            }
        }

    }
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        intervals.add(newInterval);
        List<Interval> temp = merge(intervals);
        Collections.sort(temp, new IntervalComparator());
        return temp;
    }
    public List<Interval> merge(List<Interval> intervals) {
        boolean[] x = new boolean[40000];
        boolean[] dots = new boolean[40000];
        int max = Integer.MIN_VALUE,min = Integer.MAX_VALUE;
        List<Interval> list = new ArrayList<Interval>();
        for(Interval interval : intervals) {
            for(int i=interval.start;i<interval.end;i++) {
                x[i]=true;
                if(i<min)
                    min=i;
                if(i>max)
                    max=i;
            }
        }

        int i = min;
        while(i<=max) {
            if(!x[i]) {
                i++;
                continue;
            }
            int start=i,end=start;
            while(x[end])
                end++;
            list.add(new Interval(start,end));
            i=end;

        }
        for(Interval interval : intervals) {
            int start = interval.start,end = interval.end;
            if(start==end && !dots[interval.start]) {
                dots[start] = true;
                if(start==0 && !x[0])
                    list.add(new Interval(start, end));
                else if(!x[start] && !x[start-1])
                    list.add(new Interval(start, end));
            }

        }
        return list;
    }
}