merge-intervals 合并区间

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].

 

合并有重叠的区间，且原区间序列无序

/**
 * 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 compare(const Interval &first,const Interval &second)
    {
        if(first.start==second.start)
            return first.end<second.end;
        else
            return first.start<second.start;
    }
    vector<Interval> merge(vector<Interval> &intervals) {
        vector<Interval> ret;
        int n=intervals.size();
        int pre=0, cur=0;
        sort(intervals.begin(),intervals.end(),compare);
        while(cur<n)
        {
            while(cur<n&&intervals[pre].end>=intervals[cur].start)
            {
                intervals[pre].start=min(intervals[pre].start,intervals[cur].start);
                intervals[pre].end=max(intervals[pre].end,intervals[cur].end);
                cur++;
            }
            ret.push_back(intervals[pre]);
            pre=cur;
            
        }
        return ret;
    }
};

 思路：先按照[0]来排序，然后将intervals[0]塞入res，遍历intervals，如果res.back()[1]>=intervals[i][0]，则res.back()[1]=max(res.back()[1],intervals[i][1])，因为可能会有[1,4],[2,3]合并。

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