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(),
             [](vector<int>& a, vector<int>& b) {
                 return a[0] == b[0] ? a[1] < b[1] : a[0] < b[0];
             });
        vector<vector<int>> res;
        for (auto& interval : intervals) {
            if (res.empty() || res.back()[1] < interval[0]) {
                res.push_back(interval);
            } else {
                res.back()[1] = max(interval[1], res.back()[1]);
            }
        }
        return res;
    }
};

Learned:
1: Collections framework 的用法。 Tuitoral

2: Comparator in C++

 https://www.geeksforgeeks.org/comparator-in-cpp/