# [Merge Intervals][title]

## Description

Given a collection of intervals, merge all overlapping intervals.

**Example 1:**

```
Input: [[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: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considerred overlapping.
```

**Tags:** Array, Sort


## 思路

题意是给你一组区间，让你把区间合并成没有交集的一组区间。我们可以把区间按 `start` 进行排序，然后遍历排序后的区间，如果当前的 `start` 小于前者的 `end`，那么说明这两个存在交集，我们取两者中较大的 `end` 即可；否则的话直接插入到结果序列中即可。

```java
/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        if (intervals == null || intervals.size() <= 1) return intervals;
        intervals.sort(new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                if (o1.start < o2.start) return -1;
                if (o1.start > o2.start) return 1;
                return 0;
            }
        });
        List<Interval> ans = new ArrayList<>();
        int start = intervals.get(0).start;
        int end = intervals.get(0).end;
        for (Interval interval : intervals) {
            if (interval.start <= end) {
                end = Math.max(end, interval.end);
            } else {
                ans.add(new Interval(start, end));
                start = interval.start;
                end = interval.end;
            }
        }
        ans.add(new Interval(start, end));
        return ans;
    }
}
```


## 结语

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]



[title]: https://leetcode.com/problems/merge-intervals
[ajl]: https://github.com/Blankj/awesome-java-leetcode