[LeetCode] 57. Insert Interval（Java）

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）在无重叠部分，直接给ans赋值，然后接着向后搜索。

（2）在有重叠部分：这时候我们需要得到新区间的前端和后端分别值为多少。新区见的前端值等于intervals该区间前端值和newInterval前端值中的较小者， 新区间后端值等于较大者，并不断后移，找到最大的后端值。

（3）找到新区间之后，将之后剩余的区间直接赋值给ans。

代码：

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()][]);
    }
}