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.

给定一个有重叠的区间集合，合并所有重叠的区间。先对区间以第一个元素进行排序，定义一个变量result记录合并后的区间。然后迭代这些区间，如果区间的开始值大于result的最后一个区间的结尾值，说明整个区间都在result的最后一个区间的右侧，直接添加到result。如果区间开始值小于result的最后一个区间的结尾值，则有重合需要合并，result中的最后一个区间的尾部值变为两个尾部中的最大值。

Java：

public class Solution {
public List<Interval> merge(List<Interval> intervals) {
if (intervals == null || intervals.size() <= 1) {
return intervals;
}

Collections.sort(intervals, new IntervalComparator());

List<Interval> result = new ArrayList<Interval>();
Interval last = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
Interval curt = intervals.get(i);
if (curt.start <= last.end ){
last.end = Math.max(last.end, curt.end);
}else{
result.add(last);
last = curt;
}
}

result.add(last);
return result;
}

private class IntervalComparator implements Comparator<Interval> {
public int compare(Interval a, Interval b) {
return a.start - b.start;
}
}

}

Java：

class Solution {
/**
* @param intervals, a collection of intervals
* @return: A new sorted interval list.
*/
public List<Interval> merge(List<Interval> intervals) {
List<Interval> ans = new ArrayList<>();

intervals.sort(Comparator.comparing(i -> i.start));

Interval last = null;
for (Interval item : intervals) {
if (last == null || last.end < item.start) {
ans.add(item);
last = item;
} else {
last.end = Math.max(last.end, item.end);
}
}
return ans;
}
}　　

Python:

"""
Definition of Interval.
class Interval(object):
def __init__(self, start, end):
self.start = start
self.end = end
"""

class Solution:
# @param intervals, a list of Interval
# @return a list of Interval
def merge(self, intervals):
intervals = sorted(intervals, key=lambda x: x.start)
result = []
for interval in intervals:
if len(result) == 0 or result[-1].end < interval.start:
result.append(interval)
else:
result[-1].end = max(result[-1].end, interval.end)
return result

C++：

class Solution {
public:
static bool comp(const Interval &a, const Interval &b) {
return (a.start < b.start);
}
vector<Interval> merge(vector<Interval> &intervals) {
vector<Interval> res;
if (intervals.empty()) return res;
sort(intervals.begin(), intervals.end(), comp);
res.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); ++i) {
if (res.back().end >= intervals[i].start) {
res.back().end = max(res.back().end, intervals[i].end);
} else {
res.push_back(intervals[i]);
}
}
return res;
}
};

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