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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]
Example 2:
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]
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
分析:分三步:(1)在無重疊部分,直接給ans賦值,然後接着向後搜索。
(2)在有重疊部分:這時候我們需要得到新區間的前端和後端分別值爲多少。新區見的前端值等於intervals該區間前端值和newInterval前端值中的較小者, 新區間後端值等於較大者,並不斷後移,找到最大的後端值。
(3)找到新區間之後,將之後剩餘的區間直接賦值給ans。
代碼:
class Solution {
public int[][] insert(int[][] intervals, int[] newInterval) {
List<int []> ans=new ArrayList<>();
int n=intervals.length;
int curr=0;
while(curr<n && newInterval[0]>intervals[curr][1]){
ans.add(intervals[curr]);
curr++;
}
while(curr<n && intervals[curr][0]<=newInterval[1]){
newInterval[0]=Math.min(newInterval[0],intervals[curr][0]);
newInterval[1]=Math.max(newInterval[1],intervals[curr][1]);
curr++;
}
ans.add(newInterval);
while(curr<n)
{
ans.add(intervals[curr]);
curr++;
}
return ans.toArray(new int[ans.size()][]);
}
}