LeetCode-Merge Intervals

题目要求:
给定一个interval的集合,合并重合的部分:
如:[[1,4],[2,6]] -> [[1,6]]

解题要点:

  1. 先对interval序列进行排序,排序的依据是interval.end;
  2. 从序列尾部依次往头合并
    合并的原则:
    当期interval
    待合并的Interval: ml
    判断interval ml能否合并,如能合并,将合并后生成的新的interval作为当前Interval,如不能合并,说明当前interval已经不能与序列中任何一个进行合并(不能合并说明当前interval.start>ml.end,序列是有序序列,说明当前interval不能与序列中任何一个Interval合并),当前的interval可以加入结果集。
/**
 * 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:
    auto merge_two(vector& intervals,Interval& interval,int index,bool& merged){
        auto ml = intervals[index];
        if (ml.endinterval.start? interval.start:ml.start;
            ml.end = interval.end;
        }
        return ml;
    }
    vector merge(vector& intervals) {
        vector res;
        if (intervals.size()<=1){
            res = intervals;
            return res;
        }
        sort(intervals.begin(),intervals.end(),[](const auto& r,const auto& l){return r.end < l.end;});
        auto interval = intervals[intervals.size()-1];
        int index = intervals.size()-2;
        bool merged = false;
        while(index >= 0){
            auto tmp = merge_two(intervals,interval,index,merged);
            if (merged){
                interval = tmp;
            }else{
                res.push_back(tmp);
                interval = intervals[index];
            }
            index--;
            merged=false;
            
        }
        res.push_back(interval);
        std::reverse(res.begin(),res.end());
        return res;
    }
};

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