92 lines
2.5 KiB
C++
92 lines
2.5 KiB
C++
// Source : https://oj.leetcode.com/problems/insert-interval/
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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 set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
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*
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* You may assume that the intervals were initially sorted according to their start times.
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*
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* Example 1:
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* Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
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*
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* Example 2:
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* Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
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*
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* This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
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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 factors 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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//just reuse the solution of "Merge Intervals", quite straight forward
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vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
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intervals.push_back(newInterval);
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return merge(intervals);
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}
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int main(int argc, char**argv)
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{
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Interval i1(1,2);
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Interval i2(3,5);
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Interval i3(6,7);
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Interval i4(8,10);
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Interval i5(12,16);
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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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intervals.push_back(i5);
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Interval n(4,9);
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vector<Interval> r = insert(intervals, n);
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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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