以数组 intervals 表示若干个区间的集合，其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间，并返回一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间

https://leetcode-cn.com/problems/merge-intervals/

示例1：

输入：intervals = [[1,3],[2,6],[8,10],[15,18]]
输出：[[1,6],[8,10],[15,18]]
解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]

示例2：

输入：intervals = [[1,4],[4,5]]
输出：[[1,5]]
解释：区间 [1,4] 和 [4,5] 可被视为重叠区间。

提示：

1 <= intervals.length <= 104
intervals[i].length == 2
0 <= start_i <= end_i <= 104

Java解法

思路：

• 依次遍历，判断其他区间与当前区间是否存在重叠，若存在进行合并跳出，再进行遍历
• 若不存在加入输出数组中
• 用到递归，所以空间占用并不是很高效

package sj.shimmer.algorithm.m2;

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

/**
 * Created by SJ on 2021/2/27.
 */

class D34 {
    public static void main(String[] args) {
        int[][] merge = merge(new int[][]{{1, 3}, {2, 6}, {8, 10}, {15, 18}});
        int[][] merge1 = merge(new int[][]{{1, 4}, {4, 5}});
        int[][] merge2 = merge(new int[][]{{2,3},{2,2},{3,3},{1,3},{5,7},{2,2},{4,6}});
        for (int[] ints : merge2) {
            for (int anInt : ints) {
                System.out.print(anInt);
                System.out.print(",");
            }
            System.out.println("---");
        }
    }
    public static int[][] merge(int[][] intervals) {
        List<int[]> list = new ArrayList<>();
        for (int[] interval : intervals) {
            list.add(interval);
        }
        List<int[]> merge = merge(list, 0);
        return merge.toArray(new int[merge.size()][2]);
    }
    public static List<int[]> merge(List<int[]> lists,int index) {
        if (lists != null&&lists.size()>index) {
            int[] compare = lists.get(index);
            for (int i = index+1; i < lists.size(); i++) {
                int[] interval = lists.get(i);
                //无重叠情况
                if (interval[0]>compare[1]||interval[1]<compare[0]) {
                    if (i==lists.size()-1) {
                        return merge(lists,++index);
                    }
                    continue;
                }else {
                    compare[0]=Math.min(interval[0],compare[0]);
                    compare[1]=Math.max(interval[1],compare[1]);
                    lists.remove(interval);
                    return merge(lists,index);
                }
            }
        }
        return lists;
    }
}

image

官方解

https://leetcode-cn.com/problems/merge-intervals/solution/he-bing-qu-jian-by-leetcode-solution/

1. 排序

   • 先按左侧边界做升序排序，这样可以合并的区间一定是连续的

   • 然后按区间重叠方式计算要加入的数据

     public int[][] merge(int[][] intervals) {
         if (intervals.length == 0) {
             return new int[0][2];
         }
         Arrays.sort(intervals, new Comparator<int[]>() {
             public int compare(int[] interval1, int[] interval2) {
                 return interval1[0] - interval2[0];
             }
         });
         List<int[]> merged = new ArrayList<int[]>();
         for (int i = 0; i < intervals.length; ++i) {
             int L = intervals[i][0], R = intervals[i][1];
             if (merged.size() == 0 || merged.get(merged.size() - 1)[1] < L) {
                 merged.add(new int[]{L, R});
             } else {
                 merged.get(merged.size() - 1)[1] = Math.max(merged.get(merged.size() - 1)[1], R);
             }
         }
         return merged.toArray(new int[merged.size()][]);
     }
     

     image

   • 时间复杂度：O(n log n)

   • 空间复杂度：O(log n)