Leetcode: Merge/Insert Interval

题目

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].

 

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].

 

思路:

1. 模拟题

2. Insert Interval 是 Merge Interval 的扩展, 将 newInterval 插入到 Intervals 即可转化成 Merge Interval

 

代码

bool sortInterval(const Interval &i1, const Interval &i2) {

	return i1.start < i2.start;

}

class Solution {

public:

    vector<Interval> merge(vector<Interval> &intervals) {

		vector<Interval> newVec;

		if(intervals.size() <= 0)

			return newVec;

		

		sort(intervals.begin(), intervals.end(), sortInterval);

		for(int i = 0; i < intervals.size(); i ++) {

			Interval nextInt = intervals[i];

			if(i == intervals.size()-1) {

				newVec.push_back(nextInt);

				break;

			}

			Interval tmp = intervals[i+1];

			while(tmp.start <= nextInt.end) {

				nextInt.end = max(tmp.end, nextInt.end);

				i++;

				if(i+1<intervals.size())

					tmp = intervals[i+1];

				else {

					newVec.push_back(nextInt);

					return newVec;

				}

			}

			newVec.push_back(nextInt);

		}

		return newVec;

    }

};