题目：
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].
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解题思路：
使用sort对vector<Interval>首元素进行升序排序。这里需要匿名函数的相关知识（http://www.cnblogs.com/pzhfei/archive/2013/01/14/lambda_expression.html）。
Code:
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */ class Solution { public: vector<Interval> merge(vector<Interval>& intervals) { vector<Interval> v; if(intervals.empty()) return v; sort(intervals.begin(),intervals.end(),[](Interval a, Interval b){return a.start< b.start;}); //这个地方"[]"为匿名函数。按照每个数组的首元素升序排列。 v.push_back(intervals[0]); for(int i = 1; i< intervals.size();i++) { if(v.back().end >= intervals[i].start) //这里的等号必不可少。如[1,4],[4,5]的情况 { v.back().end = max(v.back().end, intervals[i].end); } else { v.push_back(intervals[i]); } } return v; } };
