1. 区间合并

先将区间进行排序，排序完后那么，区间合并就为以下三种情况

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

 2. 无重叠区间

假设按照左边界进行排序，那么从第二个元素开始遍历，如果它的左边界在前一个元素的右边界之内那么就是发生了重叠，此时需要考虑是移除前一个元素还是移除当前元素，可以轻松得出谁的右边界更长那么更有可能与后面的元素发生重叠，也就需要被移除

class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        if (intervals.size() == 0) return 0;
        sort(intervals.begin(),intervals.end());
        for (int i=0;i<intervals.size();i++){
            cout<<intervals[i][0]<<","<<intervals[i][1]<<endl;
        }
        int st,ed;
        int result = 0;
        st = intervals[0][0];
        ed = intervals[0][1];
        for (int i=1;i<intervals.size();i++){
            if (intervals[i][0]<ed){
                result++;
                ed = min(intervals[i][1],ed);
            }else {
                ed = intervals[i][1];
            }
        }
        return result;
    }
};