Insert Interval

原创已于 2022-04-17 06:29:14 修改 · 571 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

于 2014-03-02 07:42:34 首次发布

Interval 专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个高效的算法，用于在一个已排序的非重叠区间集合中插入一个新的区间，并在必要时进行合并。通过两个实例展示了算法的具体应用过程。该算法考虑了三种情况：新区间位于当前区间之前、之后或与之相交。

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as[1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as[1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

思路：分三种情况，一种是interval 在newInterval前面，一种是newInterval在interval前面，一种是相交；记住永远是判断curInterval和newInterval的关系；

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> list = new ArrayList<>();
        for(int[] curInterval: intervals) {
            if(curInterval[1] < newInterval[0]) {
                list.add(curInterval);
            } else if(newInterval[1] < curInterval[0]) {
                list.add(newInterval);
                newInterval = curInterval;
            } else {
                newInterval[0] = Math.min(curInterval[0], newInterval[0]);
                newInterval[1] = Math.max(curInterval[1], newInterval[1]);
            }
        }
        list.add(newInterval);
        
        int[][] res = new int[list.size()][2];
        for(int i = 0; i < list.size(); i++) {
            res[i][0] = list.get(i)[0];
            res[i][1] = list.get(i)[1];
        }
        return res;
    }
}