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].

Tips:

和Merge Intervals差不多，不断比较interval的start和end，进行插入或合并，分各种情况讨论。

Code:

/**
 * 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> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> result = new ArrayList<>();
        if (intervals == null || intervals.size() == 0) {
            result.add(newInterval);
            return result;
        }
        int index = 0;
        boolean inserted = false;
        while (index < intervals.size()) {
            Interval current = intervals.get(index);
            if (newInterval.start < current.start) {
                if (newInterval.end < current.start) {
                    result.add(newInterval);
                    inserted = true;
                    break;
                }
                else {
                    if (newInterval.end <= current.end) {
                        result.add(new Interval(newInterval.start, current.end));
                        index++;
                        inserted = true;
                        break;
                    }
                    else {
                        index++;
                    }
                }
            }
            else {
                if (newInterval.start > current.end) {
                    result.add(current);
                    index++;
                }
                else {
                    if (newInterval.end <= current.end) {
                        inserted = true;
                        break;
                    }
                    else {
                        if (index == intervals.size() - 1) {
                            result.add(new Interval(current.start, newInterval.end));
                            inserted = true;
                        }
                        else {
                            newInterval.start = current.start;
                        }
                        index++;
                    }                    
                }
            }
        }
        while (index < intervals.size()) {
            result.add(intervals.get(index));
            index++;
        }
        if (!inserted) {
            result.add(newInterval);
        }
        return result;
    }

}