56. Merge Intervals

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.

难度：medium

题目：给定间隔点集合，合并所有重复的间隔。

思路：先排序然后合并。

Runtime: 57 ms, faster than 21.60% of Java online submissions for Merge Intervals.
Memory Usage: 35 MB, less than 3.23% of Java online submissions for Merge Intervals.

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        if (null == intervals || intervals.isEmpty()) {
            return new ArrayList<>();
        }
        
        Collections.sort(intervals, (o1, o2) -> {
            Interval i1 = (Interval) o1;
            Interval i2 = (Interval) o2;
            if (0 == Integer.compare(i1.start, i2.start)) {
                return Integer.compare(i1.end, i2.end);
            }
            return Integer.compare(i1.start, i2.start);
        });
        
        Stack<Interval> stack = new Stack<>();
        stack.push(intervals.get(0));
        for (int i = 1; i < intervals.size(); i++) {
            Interval i1 = stack.pop();
            Interval i2 = intervals.get(i);
            if (i1.end >= i2.start) {
                stack.push(new Interval(i1.start, Math.max(i1.end, i2.end)));
            } else {
                stack.push(i1);
                stack.push(i2);
            }
        }
        
        return new ArrayList<>(stack);
    }
}