56. Merge Intervals -M

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

A:

自己写了Comparator里的compare函数。

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

Learned:
1: Collections framework 的用法。 Tuitoral