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].

题目的意思是将相交得区间合并,典型的贪心算法

首先将区间先按照start进行排序,

然后保存先前区间的start和end

如果当前的start > 先前的end,说明当前的区间与之前的区间不想交,则将先前的区间放入结果中,同时更新start和end

如果当前的start < 先前的end,说明当前的区间与先前的区间相交,故比较当前的end与先前的end,如果当前的end大于先前的end,则更新先前的end为当前的end

bool cmp(const Interval& a, const Interval& b){
if(a.start!=b.start) return a.start < b.start;
else return a.end < b.end;
} class Solution {
public:
vector<Interval> merge(vector<Interval> &intervals) {
sort(intervals.begin(), intervals.end(), cmp);
vector<Interval> res;
if(intervals.size() == ) return res;
int start = intervals[].start, end = intervals[].end;
for(int i = ; i < intervals.size(); ++ i){
if(intervals[i].start > end){
res.push_back(Interval(start,end));
start = intervals[i].start, end = intervals[i].end;
}else{
end = max(end, intervals[i].end);
}
}
res.push_back(Interval(start,end));
return res;
}
};
 

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