Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).java
You may assume that the intervals were initially sorted according to their start times.数组
Example 1:app
Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]
Example 2:code
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8] Output: [[1,2],[3,10],[12,16]] Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
给定一个左端点递增且互不重叠的区间数组,向其中插入一个新的区间,若有重叠则须要合并,返回处理后的区间数组。it
先将全部右端点在新区间左端点以前的区间(即必定不会相加)插入到结果中,再将全部能够与新区间合并的区间进行合并,并加入到结果中,最后将剩余一部分不会合并的区间加入到结果中。io
class Solution { public int[][] insert(int[][] intervals, int[] newInterval) { List<int[]> list = new ArrayList<>(); int i = 0; while (i < intervals.length && intervals[i][1] < newInterval[0]) { list.add(intervals[i++]); } while (i < intervals.length && intervals[i][0] <= newInterval[1] && intervals[i][1] >= newInterval[0]) { newInterval[0] = Math.min(newInterval[0], intervals[i][0]); newInterval[1] = Math.max(newInterval[1], intervals[i][1]); i++; } list.add(newInterval); while (i < intervals.length) { list.add(intervals[i++]); } return list.toArray(new int[0][]); } }