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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]
.
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */ 10 11 bool cmp(const Interval &a, const Interval &b) { 12 return a.start < b.start; 13 } 14 15 class Solution { 16 public: 17 vector<Interval> merge(vector<Interval> &intervals) { 18 int pos = 0, cnt = 0; 19 sort(intervals.begin(), intervals.end(), cmp); 20 for (int i = 1; i < intervals.size(); ++i) { 21 if (intervals[pos].end >= intervals[i].start) { 22 ++cnt; 23 if (intervals[pos].end <= intervals[i].end) { 24 intervals[pos].end = intervals[i].end; 25 } 26 } else { 27 ++pos; 28 intervals[pos].start = intervals[i].start; 29 intervals[pos].end = intervals[i].end; 30 } 31 } 32 intervals.resize(intervals.size()-cnt); 33 return intervals; 34 } 35 };
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原文地址:http://www.cnblogs.com/easonliu/p/4215802.html