[LeetCode] Merge Intervals_综合

[Problem]

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].

[Analysis]
唯一需要注意的时，cmp函数需要写在Solution类外面。

[Solution]

//
// Definition for an interval.
// struct Interval {
      
//     int start;
//     int end;
//     Interval() : start(0), end(0) {}
//     Interval(int s, int e) : start(s), end(e) {}
// };
//
// compare
bool cmp(const Interval &inte1, const Interval &inte2){
      
    return inte1.start < inte2.start;
}

// definition of Solution
class Solution {
      
public:
    // check the two intervals wether they are interlocked
    bool check(Interval inter1, Interval inter2){
      
        if((inter1.start >= inter2.start && inter1.start <= inter2.end) || (inter1.end >= inter2.start && inter1.end <= inter2.end)){
      
            return true;
        }
        else if((inter2.start >= inter1.start && inter2.start <= inter1.end) || (inter2.end >= inter1.start && inter2.end <= inter1.end)){
      
            return true;
        }
        else{
      
            return false;
        }
    }
   
    // merge the two intervals
    Interval merge(Interval inter1, Interval inter2){
      
        Interval inter;
        inter.start = inter1.start < inter2.start ? inter1.start : inter2.start;
        inter.end = inter1.end > inter2.end ? inter1.end : inter2.end;
        return inter;
    }
   
    // merge
    vector<Interval> merge(vector<Interval> &intervals) {
      
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> res;
   
        // sort
        sort(intervals.begin(), intervals.end(), cmp);
        for(int i = 0; i < intervals.size(); ++i){
      
            Interval inte = intervals[i];
            if(res.size() > 0 && check(res.back(), inte)){
      
                while(res.size() > 0 && check(res.back(), inte)){
      
                    inte = merge(res.back(), intervals[i]);
                    res.pop_back();
                }
            }
            res.push_back(inte);
        }
        return res;
    }
};