Leetcode | Merge Intervals

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].

这里主要注意的是STL的sort的函数的comparor要写成static函数，而且参数采用值传递，不要用引用。

class Solution {

public:

    static bool comp(Interval m1, Interval m2) {

        return m1.start < m2.start;

    }



    vector<Interval> merge(vector<Interval> &intervals) {

        sort(intervals.begin(), intervals.end(), Solution::comp);

        vector<Interval> ret;

        if (intervals.empty()) return ret;

        Interval pre(intervals[0]);

        for (int i = 0; i < intervals.size(); ++i) {

            if (intervals[i].start <= pre.end) {

                if (intervals[i].end > pre.end) pre.end = intervals[i].end;

            } else {

                ret.push_back(pre);

                pre = intervals[i];

            }

        }

        ret.push_back(pre);

        

        return ret;

    }

};

 这样写。

 1 /**

 2  * Definition for an interval.

 3  * struct Interval {

 4  *     int start;

 5  *     int end;

 6  *     Interval() : start(0), end(0) {}

 7  *     Interval(int s, int e) : start(s), end(e) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     static bool compare(const Interval &i1, const Interval &i2) {

13         return i1.start < i2.start;

14     }

15     vector<Interval> merge(vector<Interval> &intervals) {

16         sort(intervals.begin(), intervals.end(), compare);

17         vector<Interval> ans;

18 

19         int n = intervals.size();

20         for (int i = 0; i < n; ++i) {

21             if (i == n - 1 || intervals[i].end < intervals[i + 1].start) {

22                 ans.push_back(intervals[i]);

23             } else {

24                 intervals[i + 1].start = intervals[i].start;

25                 if (intervals[i].end > intervals[i + 1].end) intervals[i + 1].end = intervals[i].end;

26             }

27         }

28         

29         return ans;

30     }

31 };