刷题56. Merge Intervals

一、题目说明

题目是56. Merge Intervals，给定一列区间的集合，归并重叠区域。

二、我的做法

这个题目不难，先对intervals排序，然后取下一个集合，如果cur[0]>resLast[1]在直接放到集合中，否者合并。代码如下：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
class Solution{
	public:
		vector<vector<int>> merge(vector<vector<int>>& intervals){
			int len = intervals.size();
			if(len<2) return intervals;
			
			sort(intervals.begin(),intervals.end());
			vector<vector<int>> res;
			
			res.push_back(intervals[0]);
			vector<int> cur,resLast;

			for(int i=1;i<len;i++){
				cur = intervals[i];
				resLast = res[res.size()-1];
					
				if(cur[0]>resLast[1]){
					res.push_back(cur);
				}else if(cur[0]<=resLast[1] && cur[1]>resLast[1]){
					res.back() = {resLast[0],cur[1]};
				}		
			}
			
			return res;
		}
};
int main(){
	Solution s;
	vector<vector<int>> m;
	vector<vector<int>> r;
	
	
	m = {{1,3},{2,6},{8,10},{15,18}};
	r = s.merge(m);
	for(int i=0;i<r.size();i++){
		for(int j=0;j<r[i].size();j++){
			cout<<r[i][j]<<"->";
		}
		cout<<"\n";
	}
	
	cout<<"--------"<<"\n";
	
	m = {{1,4},{4,5}};
	r = s.merge(m);
	for(int i=0;i<r.size();i++){
		for(int j=0;j<r[i].size();j++){
			cout<<r[i][j]<<"->";
		}
		cout<<"\n";
	}
	
	m = {{1,4},{0,4}};
	r = s.merge(m);
	for(int i=0;i<r.size();i++){
		for(int j=0;j<r[i].size();j++){
			cout<<r[i][j]<<"->";
		}
		cout<<"\n";
	}
	
	m = {{1,4},{2,3}};
	r = s.merge(m);
	for(int i=0;i<r.size();i++){
		for(int j=0;j<r[i].size();j++){
			cout<<r[i][j]<<"->";
		}
		cout<<"\n";
	}	
	return 0;
}

性能如下：

Runtime: 24 ms, faster than 48.18% of C++ online submissions for Merge Intervals.
Memory Usage: 12.5 MB, less than 82.56% of C++ online submissions for Merge Intervals.

三、优化措施

暂时这样，不优化了。