Insert Interval

问题描述

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].

解决思路

分三部分：

1. 把之前的较小区间加入list；

2. 更新重叠的区间的start和end，并加入新的区间；

3. 将剩余的区间加入。

程序

public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        if (newInterval == null) {
            return intervals;
        }
        List<Interval> merge = new ArrayList<Interval>();
        if (intervals.size() == 0) {
            merge.add(newInterval);
            return merge;
        }
        
        int i = 0, n = intervals.size();
        // add smaller
        while (i < n && intervals.get(i).end < newInterval.start) {
            merge.add(intervals.get(i++));
        }
        // merge
        while (i < n && intervals.get(i).start <= newInterval.end) {
            newInterval.start = Math.min(newInterval.start, intervals.get(i).start);
            newInterval.end = Math.max(newInterval.end, intervals.get(i).end);
            ++i;
        }
        merge.add(newInterval);
        // add last
        while (i < n) {
            merge.add(intervals.get(i++));
        }
        return merge;
    }
}

Follow up

Merge Intervals

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]. 

解决思路

1. 排序，从小到大；

2. merge。

程序

public class Solution {
    public List<Interval> merge(List<Interval> intervals) {
		List<Interval> mergeList = new ArrayList<Interval>();
		if (intervals == null || intervals.isEmpty()) {
			return mergeList;
		}
		
		sortedIntervalsList(intervals);
		mergeList.add(intervals.get(0));
		
		for (int i = 1; i < intervals.size(); i++) {
			Interval last = mergeList.get(mergeList.size() - 1);
			Interval insert = intervals.get(i);
			if (insert.start <= last.end) {
				last.end = Math.max(last.end, insert.end);
			} else {
				mergeList.add(insert);
			}
		}
		
		return mergeList;
	}

	private void sortedIntervalsList(List<Interval> intervals) {
		Comparator<Interval> comp = new Comparator<Interval>() {
			@Override
			public int compare(Interval i1, Interval i2) {
				if (i1.start > i2.start) {
					return 1;
				} else if (i1.start < i2.start) {
					return -1;
				}
				if (i1.end > i2.end) {
					return 1;
				} else if (i1.end < i2.end) {
					return -1;
				}
				return 0;
			}
		};
		Collections.sort(intervals, comp);
	}
}