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].
题意：去掉重复的区间。
思路：排序后，再比较end的情况。
/** * Definition for an interval. * struct Interval { *int start; *int end; *Interval() : start(0), end(0) {} *Interval(int s, int e) : start(s), end(e) {} * }; */ bool cmp(const Interval &a, const Interval &b) {return a.start < b.start; }class Solution {public:vector<Interval> merge(vector<Interval> &intervals) {int n = intervals.size();vector<Interval> ans;sort(intervals.begin(), intervals.end(), cmp);for (int i = 0; i < intervals.size(); i++) {if (ans.size() == 0) ans.push_back(intervals[i]);else {int m = ans.size();if (ans[m-1].start <= intervals[i].start && ans[m-1].end >= intervals[i].start)ans[m-1].end = max(ans[m-1].end, intervals[i].end);else ans.push_back(intervals[i]);}}return ans;}};
，经受雨，面对另一个轮回。
