以數(shù)組 intervals 表示若干個(gè)區(qū)間的集合，其中單個(gè)區(qū)間為 intervals[i] = [starti, endi] 泞边。請你合并所有重疊的區(qū)間袖肥，并返回一個(gè)不重疊的區(qū)間數(shù)組障本，該數(shù)組需恰好覆蓋輸入中的所有區(qū)間

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

示例1：

輸入：intervals = [[1,3],[2,6],[8,10],[15,18]]
輸出：[[1,6],[8,10],[15,18]]
解釋：區(qū)間 [1,3] 和 [2,6] 重疊, 將它們合并為 [1,6]

示例2：

輸入：intervals = [[1,4],[4,5]]
輸出：[[1,5]]
解釋：區(qū)間 [1,4] 和 [4,5] 可被視為重疊區(qū)間员凝。

提示：

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

Java解法

思路：

• 依次遍歷，判斷其他區(qū)間與當(dāng)前區(qū)間是否存在重疊瞧掺，若存在進(jìn)行合并跳出耕餐，再進(jìn)行遍歷
• 若不存在加入輸出數(shù)組中
• 用到遞歸，所以空間占用并不是很高效

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);
                //無重疊情況
                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

   • 先按左側(cè)邊界做升序排序辟狈，這樣可以合并的區(qū)間一定是連續(xù)的

   • 然后按區(qū)間重疊方式計(jì)算要加入的數(shù)據(jù)

     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

   • 時(shí)間復(fù)雜度：O(n log n)

   • 空間復(fù)雜度：O(log n)