76 lines
1.9 KiB
C++
76 lines
1.9 KiB
C++
// Source : https://oj.leetcode.com/problems/merge-intervals/
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// Author : Hao Chen
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// Date : 2014-08-26
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/**********************************************************************************
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*
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* Given a collection of intervals, merge all overlapping intervals.
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*
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* For example,
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* Given [1,3],[2,6],[8,10],[15,18],
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* return [1,6],[8,10],[15,18].
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*
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*
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**********************************************************************************/
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#include <iostream>
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#include <vector>
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#include <algorithm>
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using namespace std;
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struct Interval {
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int start;
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int end;
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Interval() : start(0), end(0) {}
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Interval(int s, int e) : start(s), end(e) {}
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};
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//Two factos sorting [start:end]
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bool compare(const Interval& lhs, const Interval& rhs){
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return (lhs.start==rhs.start) ? lhs.end < rhs.end : lhs.start < rhs.start;
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}
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vector<Interval> merge(vector<Interval> &intervals) {
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vector<Interval> result;
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if (intervals.size() <= 0) return result;
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//sort the inervals. Note: using the customized comparing function.
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sort(intervals.begin(), intervals.end(), compare);
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for(int i=0; i<intervals.size(); i++) {
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int size = result.size();
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// if the current intervals[i] is overlapped with previous interval.
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// merge them together
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if( size>0 && result[size-1].end >= intervals[i].start) {
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result[size-1].end = max(result[size-1].end, intervals[i].end);
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}else{
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result.push_back(intervals[i]);
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}
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}
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return result;
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}
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int main(int argc, char**argv)
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{
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Interval i1(1,4);
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Interval i2(0,2);
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Interval i3(3,5);
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Interval i4(15,18);
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vector<Interval> intervals;
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intervals.push_back(i1);
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intervals.push_back(i2);
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intervals.push_back(i3);
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intervals.push_back(i4);
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vector<Interval> r = merge(intervals);
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for(int i=0; i<r.size(); i++){
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cout << "[ " << r[i].start << ", " << r[i].end << " ] ";
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}
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cout <<endl;
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return 0;
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}
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