题目描述
Given a collection of intervals, merge all overlapping intervals.
例子
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.
代码
Java
class Solution {
public int[][] merge(int[][] intervals) {
if ( intervals.length < 2 ) return intervals;
Arrays.sort(intervals, new Comparator<int[]>(){
public int compare(int[] a, int[] b){
return a[0]-b[0];
}
});
List<int[]> list = new ArrayList<>();
int[] dummy = intervals[0];
list.add(dummy);
for(int[] interval : intervals){
if (interval[0] <= dummy[1]){
dummy[1] = Math.max(interval[1], dummy[1]);
}
else {
dummy = interval;
list.add(dummy);
}
}
return list.toArray(new int[list.size()][]);
}
}
Python3
class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
ans = []
for interval in sorted(intervals, key = lambda x:x[0]):
if not ans or interval[0] > ans[-1][1]:
ans.append(interval)
else :
ans[-1][1] = max(ans[-1][1], interval[1])
return ans