https://leetcode.com/problems/insert-interval/description/
You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.
Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).
Return intervals after the insertion.
Example 1:
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<strong>Input:</strong> intervals = [[1,3],[6,9]], newInterval = [2,5] <strong>Output:</strong> [[1,5],[6,9]] |
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class Solution { public int[][] insert(int[][] intervals, int[] newInterval) { List<int[]> res = new ArrayList<>(); //add newInterval to the list res.add(newInterval); //add intervals to the list for(int[] i: intervals){ res.add(i); } //sort the list by ascending order by start Collections.sort(res, new Comparator<int[]>(){ public int compare(int[] a, int[] b){ return Integer.compare(a[0], b[0]); } }); int[] prev = res.get(0); List<int[]> list = new ArrayList<>(); int i=1; while(i<res.size()){ //overlapping next interval if (prev[0] <= res.get(i)[0] && res.get(i)[0] <= prev[1]){ prev[0] = Math.min(prev[0], res.get(i)[0]); prev[1] = Math.max(prev[1], res.get(i)[1]); i++; } //if not overalapping else { list.add(prev); prev = res.get(i); i++; } } list.add(prev); //final ans int[][] ans = new int[list.size()][2]; int j=0; for(int[] r: list){ ans[j++] = r; } return ans; } } |