Insert Interval 30

Question

Given a non-overlapping interval list which is sorted by start point.

Insert a new interval into it, make sure the list is still in order and non-overlapping (merge intervals if necessary).

Example

Insert [2, 5] into [[1,2], [5,9]], we get [[1,9]].

Insert [3, 4] into [[1,2], [5,9]], we get [[1,2], [3,4], [5,9]].

Solution

用pos记录newInterval应该插入的位置。顺序遍历intervals中的元素，若当前interval的end比newInterval的start还小，则将当前interval加入答案，同时pos＋1；若比newInterval大，则直接加入答案；若有overlap，则需要merge，newInterval的start取两者间小的，end取两者间大的。最后在pos的位置插入newInterval即可。

代码如下：

/**
 * Definition of Interval:
 * public classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this.start = start;
 *         this.end = end;
 *     }
 */

class Solution {
    /**
     * Insert newInterval into intervals.
     * @param intervals: Sorted interval list.
     * @param newInterval: A new interval.
     * @return: A new sorted interval list.
     */
    public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
        ArrayList<Interval> result = new ArrayList<Interval>();
        // write your code here
        if(newInterval == null){
            result = intervals;
            return result;
        }

        if(intervals == null || intervals.size() == 0){
            result.add(newInterval);
            return result;
        }

        int pos = 0;
        for(Interval interval : intervals){
            if(interval.end < newInterval.start){
                result.add(interval);
                pos++;
            }else if(interval.start > newInterval.end){
                result.add(interval);
            }else{
                newInterval.start = Math.min(interval.start, newInterval.start);
                newInterval.end = Math.max(interval.end, newInterval.end);
            }
        }

        result.add(pos, newInterval);

        return result;
    }
}