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
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