Leetcode 56. Merge Intervals

56. Merge Intervals

• Total Accepted: 75204
• Total Submissions: 285007
• Difficulty: Hard

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].

思路：先根据区间左边界大小升序排序区间，然后向右合并。注意sort默认是升序排序，

代码：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    static bool comp(Interval a,Interval b){
        return a.start<b.start;
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        vector<Interval> res;
        if(!intervals.size()) return res;
        sort(intervals.begin(),intervals.end(),comp);
        res.push_back(intervals[0]);
        for(int i=1;i<intervals.size();i++){
            if(intervals[i].start>res.back().end){
                res.push_back(intervals[i]);
            }
            else{
                res.back().end=max(intervals[i].end,res.back().end);
            }
        }
        return res;
    }
};

sort排序也可以这样写：

class Solution {
public:
    vector<Interval> merge(vector<Interval>& intervals) {
        vector<Interval> res;
        if(!intervals.size()) return res;
        sort(intervals.begin(),intervals.end(),[](Interval a, Interval b){return a.start < b.start;});
        res.push_back(intervals[0]);
        for(int i=1;i<intervals.size();i++){
            if(intervals[i].start>res.back().end){
                res.push_back(intervals[i]);
            }
            else{
                res.back().end=max(intervals[i].end,res.back().end);
            }
        }
        return res;
    }
};