[LeetCode] 56. Merge Interval
Posted 蘑菇君520
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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 considered overlapping.
Solution
思路还是很清楚的,按start从小到大排序,然后遍历整个list,遇到两个重叠的就取更大的end。但是自己实现的不够优雅,在循环外对第一个和最后一个数据做了额外处理。
public Interval merge(List<Interval> intervals)
if (intervals == null) return null;
if (intervals.length == 0) return new ArrayList<>();
Collections.sort(intervals, new Comparator<Interval>()
public int compare(Interval o1, Interval o2)
return o1.start - o2.start;
);
// 先拿出第一个数据做初始值
int start = intervals.get(0).start;
int end = intervals.get(0).end;
for (int i=1; i<intervals.size(); i++)
Interval interval = intervals.get(i);
if (end >= interval.start)
end = Math.max(end, interval.end);
else
res.add(new Interval(start, end));
start = interval.start;
end = interval.end;
// 这里要对最后一个Interval做额外处理
res.add(new Interval(start, end));
return res;
然后看别人的答案,贪吃蛇的写法,甚是优雅~
private class IntervalComparator implements Comparator<Interval>
@Override
public int compare(Interval a, Interval b)
return a.start - b.start ;
public List<Interval> merge(List<Interval> intervals)
Collections.sort(intervals, new IntervalComparator());
LinkedList<Interval> merged = new LinkedList<Interval>();
for (Interval interval : intervals)
// if the list of merged intervals is empty or if the current
// interval does not overlap with the previous, simply append it.
if (merged.isEmpty() || merged.getLast().end < interval.start)
merged.add(interval);
// otherwise, there is overlap, so we merge the current and previous
// intervals.
else
merged.getLast().end = Math.max(merged.getLast().end, interval.end);
return merged;
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