Leetcode56. consolidation interval

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One 、 Ideas

adopt drawing Looking for a regular Interval problem first Sort

Two 、 problem

In array intervals Represents a set of intervals , The single interval is intervals[i] = [starti, endi] . Please merge all overlapping intervals , And return a non overlapping interval array , The array needs to cover exactly all the intervals in the input .

Example 1：

 Input ：intervals = [[1,3],[2,6],[8,10],[15,18]]
 Output ：[[1,6],[8,10],[15,18]]
 explain ： Section  [1,3]  and  [2,6]  overlap ,  Combine them into  [1,6].

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Example  2：

 Input ：intervals = [[1,4],[4,5]]
 Output ：[[1,5]]
 explain ： Section  [1,4]  and  [4,5]  Can be regarded as overlapping interval .

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Tips ：

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Related Topics

• Sort
• Array

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3、 ... and 、 Code

public int[][] merge(int[][] intervals) {
            Arrays.sort(intervals, (o1, o2) -> {
                return o1[0] == o2[0] ? o2[1] - o1[1] : o1[0] - o2[0];
            });
            int start = intervals[0][0];
            int end = intervals[0][1];
            int[][] result = new int[intervals.length][2];
            ArrayList<int[]> ans = new ArrayList<>();
            for (int i = 0; i < intervals.length; i++) {
                if (intervals[i][0] > end) {
                    ans.add(new int[]{start, end});
                    start = intervals[i][0];
                    end = intervals[i][1];
                } else if (intervals[i][1] > end) {
                    end = intervals[i][1];
                }
            }
            ans.add(new int[]{start, end});
            return ans.toArray(new int[ans.size()][]);
        }

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