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LeetCode - Merge Intervals

时间:2021-02-13 08:46:50      阅读:27      评论:0      收藏:0      [点我收藏+]
Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

 

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
 

Constraints:

1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104
class Solution {
    public int[][] merge(int[][] intervals) {
        // sort intervals array by their start
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        List<int[]> res = new ArrayList<>();
        for (int i = 0; i < intervals.length - 1; i++) {
            if (intervals[i][1] < intervals[i+1][0]) {
                res.add(intervals[i]);
            }
            else {
                intervals[i+1][0] = intervals[i][0];
                intervals[i+1][1] = Math.max(intervals[i][1], intervals[i+1][1]);
            }
        }
        res.add(intervals[intervals.length - 1]);
        return res.toArray(new int[res.size()][2]);
    }
}

 

LeetCode - Merge Intervals

原文:https://www.cnblogs.com/incrediblechangshuo/p/14398902.html

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