package com.bupt;import java.util.ArrayList;import java.util.List;//Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).////You may assume that the intervals were initially sorted according to their start times.////Example 1://Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].////Example 2://Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].////This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].class Interval { int start; int end; Interval() { start = 0; end = 0; } Interval(int s, int e) { start = s; end = e; }}public class Solution { public static List<Interval> insert(List<Interval> intervals, Interval newInterval) { List<Interval> result = new ArrayList<Interval>(); while(intervals.size() == 0){ result.add(newInterval); return result; } int i = 0; while(i<intervals.size() && newInterval.start>intervals.get(i).end){ result.add(intervals.get(i)); i++; } if(i<intervals.size()){ newInterval.start = Math.min(newInterval.start, intervals.get(i).start); } result.add(newInterval); while(i<intervals.size() && newInterval.end>=intervals.get(i).start){ newInterval.end = Math.max(newInterval.end, intervals.get(i).end); i++; } while(i<intervals.size()){ result.add(intervals.get(i)); i++; } return result; } public static void main(String[] args){ Interval interval1 = new Interval(1,5);// Interval interval2 = new Interval(10,11);// Interval interval3 = new Interval(15,16);// Interval interval4 = new Interval(9,11); Interval newInterval = new Interval(0,1); List<Interval> intervals = new ArrayList<Interval>(); intervals.add(interval1);// intervals.add(interval2);// intervals.add(interval3);// intervals.add(interval4); List<Interval> result = insert(intervals,newInterval); for(int i = 0;i<result.size();i++){ System.out.PRintln(result.get(i).start+" "+result.get(i).end); } }}
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