[LC] 57. Insert Interval

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].

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]);
            i += 1;
        }
        int start = newInterval[0], end = newInterval[1];
        // should be start <= newInterval end
        while(i < intervals.length && intervals[i][0] <= newInterval[1]) {
            start = Math.min(start, intervals[i][0]);
            end = Math.max(end, intervals[i][1]);
            i += 1;
        }
        list.add(new int[]{start, end});
        while (i < intervals.length) {
            list.add(intervals[i]);
            i += 1;
        }
        int[][] res = new int[list.size()][2];
        for (int j = 0; j < list.size(); j++) {
            res[j] = list.get(j);
        }
        return res;
    }
}
原文地址:https://www.cnblogs.com/xuanlu/p/13128865.html