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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0056_merge_intervals;

// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Sorting
// #Data_Structure_II_Day_2_Array #Level_2_Day_17_Interval #Udemy_2D_Arrays/Matrix
// #Big_O_Time_O(n_log_n)_Space_O(n) #2023_08_11_Time_8_ms_(96.27%)_Space_45.2_MB_(90.13%)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * 56 - Merge Intervals\.
 *
 * Medium
 *
 * Given an array of `intervals` where intervals[i] = [start_i, end_i], 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 <= start_i <= end_i <= 10⁴
**/
public class Solution {
    public int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        List list = new ArrayList<>();
        int[] current = intervals[0];
        list.add(current);
        for (int[] next : intervals) {
            if (current[1] >= next[0]) {
                current[1] = Math.max(current[1], next[1]);
            } else {
                current = next;
                list.add(current);
            }
        }
        return list.toArray(new int[list.size()][]);
    }
}