[LeetCode]Insert Interval

最新推荐文章于 2017-01-10 06:40:16 发布

转载最新推荐文章于 2017-01-10 06:40:16 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/Rosanna/p/3483844.html

本文介绍了一种在已排序的非重叠区间中插入新区间并进行必要合并的方法。通过两个示例说明了如何实现这一过程，特别是当新区间与已有区间重叠时的处理方式。该算法适用于需要管理和更新一系列区间的场景。

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].

思考：与Merge Interval一样。

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
 bool comp(const Interval &a,const Interval &b)
{
    return a.start<b.start;
}
class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
		vector<Interval> ret;
		intervals.push_back(newInterval);
		sort(intervals.begin(),intervals.end(),comp);
		int len=intervals.size();
		if(len==1) return intervals;
		int left=intervals[0].start;
		int right=intervals[0].end;
		for(int i=1;i<len;i++)
		{
			if(intervals[i].start<=right)
				right=max(right,intervals[i].end);
			if(intervals[i].start>right)
			{
				ret.push_back(Interval(left,right));
				left=intervals[i].start;
				right=intervals[i].end;
			}
			if(i==len-1)
			{
		    	ret.push_back(Interval(left,right));
			}
		}
		return ret;
    }
};

转载于:https://www.cnblogs.com/Rosanna/p/3483844.html