57. Insert Interval

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].
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这是要实现区间的并集，思路是，遍历已经存在的区间，当新区间end<遍历元素的start,遍历结束，其余的有如下几种情况：1，新区间的start在当前的区间之间，但是end在当前元素后面，则将新区间的start=当前元素的start；2，新区间的end在当前区间内，但是start在区间前面，则将新区间end=当前区间的end；3，当前区间包括新区间，则令新区间=当前区间；4，新区间包括当前区间，则进行下一个区间的比较。当跳出循环之后，将新区间放入list中，还有原list中没有比较的区间。具体代码如下：

public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
        ArrayList<Interval>newIntervals=new ArrayList<>();
        int i=0;
        for (; i < intervals.size(); i++) {
            Interval interval=intervals.get(i);
            if (interval.start>newInterval.end) break;
            if (newInterval.start>=interval.start&&newInterval.start<=interval.end&&newInterval.end>interval.end) 
                newInterval.start=interval.start;
            if (newInterval.start<interval.start&&newInterval.end>=interval.start&&newInterval.end<=interval.end)
                newInterval.end=interval.end;
            if (newInterval.start>=interval.start&&newInterval.end<=interval.end){
                newInterval.start=interval.start;
                newInterval.end=interval.end;
            }
            if (newInterval.start<=interval.start&&newInterval.end>=interval.end) continue;
            if (newInterval.start>interval.end) newIntervals.add(interval);         
        }
        newIntervals.add(newInterval);
        for (; i < intervals.size(); i++) {
            newIntervals.add(intervals.get(i));
        }       
        return newIntervals;
    }