Insert Intervals

最新推荐文章于 2020-05-06 00:53:04 发布

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本文介绍了一个高效的算法，用于在一个已排序的非重叠区间集合中插入一个新的区间，并在必要时进行合并。通过两个实例展示了算法如何工作：第一个例子将[2,5]插入到[1,3],[6,9]中得到[1,5],[6,9]；第二个例子将[4,9]插入到[1,2],[3,5],[6,7],[8,10],[12,16]中得到[1,2],[3,10],[12,16]。该算法确保了新插入的区间与已有区间正确地合并。

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].

/**
 * 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 ArrayList<Interval> insert(ArrayList<Interval> intervals,
			Interval newInterval) {
		// Start typing your Java solution below
		// DO NOT write main() function
		int size = intervals.size();
		ArrayList<Interval> res = new ArrayList<Interval>();
		for(int i = 0; i < size; i++){
			Interval cur = intervals.get(i);
			if(cur.end < newInterval.start)
				res.add(new Interval(cur.start, cur.end));
			else if(cur.end >= newInterval.start && cur.start <= newInterval.end){
				newInterval.start = Math.min(newInterval.start, cur.start);
				newInterval.end = Math.max(newInterval.end, cur.end);
			}
			else{// cur.start > newInterval.end
				res.add(newInterval);
				while(i < size){
					res.add(new Interval(intervals.get(i).start, intervals.get(i).end));
					i++;
				}
			}
		}
		if(res.size() == 0 || res.get(res.size() - 1).end < newInterval.start)
			res.add(newInterval);
		return res;
	}
}