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For interval list [[1,10],[2,3],[5,8],[4,7]], return 3
If landing and flying happens at the same time, we consider landing should happen at first.
http://www.lintcode.com/en/problem/number-of-airplanes-in-the-sky/
将所有区间的start与end放在一起排序，但是要标记其是属性，然后统一排序，问题就转化成了括号匹配的问题了，只要用一个cnt来记录，遇到start就加1，遇到end就减1，记录过程中的最大值就是答案。
1 /** 2 * Definition of Interval: 3 * classs Interval { 4 * int start, end; 5 * Interval(int start, int end) { 6 * this->start = start; 7 * this->end = end; 8 * } 9 */ 10 class Solution { 11 public: 12 /** 13 * @param intervals: An interval array 14 * @return: Count of airplanes are in the sky. 15 */ 16 int countOfAirplanes(vector<Interval> &airplanes) { 17 // write your code here 18 vector<pair<int, bool> > v; 19 for (auto &i : airplanes) { 20 v.push_back(make_pair(i.start, true)); 21 v.push_back(make_pair(i.end, false)); 22 } 23 int cnt = 0, res = 0; 24 sort(v.begin(), v.end()); 25 for (auto &i : v) { 26 if (i.second) ++cnt; 27 else --cnt; 28 res = max(cnt, res); 29 } 30 return res; 31 } 32 };
[LintCode] Number of Airplanes in the Sky
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原文地址：http://www.cnblogs.com/easonliu/p/4504647.html
