算法-leetcode-排序问题-56. 合并区间

26, 56. 合并区间

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

思路：就把开始数据排序，然后判断是否能合并即可

package com.shangguigu.dachang.algrithm.A08_sort;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;

/**
 * @author : 不二
 * @date : 2022/4/20-上午10:49
 * @desc : 56. 合并区间
 * https://leetcode-cn.com/problems/merge-intervals/
 *
 **/
public class A88_merge {

    public static void main(String[] args) {
        int[][] intervals = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};
        int[][] merge = merge(intervals);
        System.out.printf(Arrays.toString(intervals));

    }

    public static int[][] merge(int[][] intervals) {

        // int[][] temp = new int[intervals.length][intervals[0].length];
        ArrayList<int[]> result = new ArrayList<>();

        // 先按照每个数组的开始数字进行排序
        Arrays.sort(intervals, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                //  从小到大
                return o1[0] - o2[0];
            }
        });

        for (int[] interval : intervals) {
            int start = interval[0];
            int end = interval[1];

            // 如果为空，直接加进去
            // 获取取最后一个区间，当前的开始值大于最后一个结束，直接加进去
            if (result.size() == 0 || start > result.get(result.size() - 1)[1]) {
                result.add(interval);
            } else {
                // 说明当前interval和之前记录的区间可以合并
                int[] lst = result.get(result.size() - 1);
                int mergeStart = lst[0];
                int mergeEnd = Math.max(lst[1], end);
                result.set(result.size() - 1, new int[]{mergeStart, mergeEnd});

            }
        }

        return result.toArray(new int[result.size()][]);
    }
}