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56.Merge-Intervals

56. Merge Intervals

题目地址

题目描述

Given a collection of intervals, merge all overlapping intervals.
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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:
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Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

代码

Approach #2 Sorting

class Solution {
public int[][] merge(int[][] intervals) {
if (intervals.length <= 1) return intervals;
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// Sort by ascending starting point
Arrays.sort(intervals, (i1, i2) -> i1[0] - i2[0]);
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List<int[]> result = new ArrayList<>();
int[] prevInterval = intervals[0];
result.add(prevInterval);
for (int[] interval : intervals) {
if (interval[0] <= prevInterval[1]) {
// Overlapping intervals, move the end if needed
prevInterval[1] = Math.max(prevInterval[1], interval[1]);
} else {
// Disjoint intervals, add the new interval to the list
prevInterval = interval;
result.add(prevInterval);
}
}
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return result.toArray(new int[result.size()][]);
}
}