题目描述

Given a collection of intervals, merge all overlapping intervals.

样例

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.

算法1

Sort then merge, $O(nlogn)$

按照start排序，然后 检查 当前的 interval 是否与 res 的last interval有overlap，如有overlap，则merge and extend last interval，否则将当前interval加到res中。

C++ 代码

    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        int n = intervals.size();
        if(n==0) return intervals;

        sort(begin(intervals), end(intervals), 
             [](const vector<int>& a, const vector<int>& b){ return a[0] < b[0]; });

        vector<vector<int>> res;
        res.push_back(intervals[0]);

        for(int i=1; i<n; i++) {
            if(intervals[i][0] <= res.back()[1]) {
                // overlap, extend the last interval in res
                res.back()[1] = max( res.back()[1], intervals[i][1] );
            } else {
                // not overlap, append new interval
                res.push_back(intervals[i]);
            }
        }

        return res;
    }