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 considerred overlapping.

思路是，先按interval的左端（start）排序。然后一个个检查是否能合并。下面这个做 in place 的替换。extra space O（1）

class Solution(object):
    def merge(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[Interval]
        """
        if not intervals:
            return []
        intervals.sort(key=lambda data: data.start)
        storeIdx = 0
        lo, hi = intervals[0].start, intervals[0].end
        for i in xrange(1, len(intervals)):
            if intervals[i].start > hi:
                intervals[storeIdx] = Interval(lo, hi)
                storeIdx += 1
                lo, hi = intervals[i].start, intervals[i].end
            else:
                hi = max(hi, intervals[i].end)

        intervals[storeIdx] = Interval(lo, hi)   # 这行不能漏。从for循环出来的时候，还有一个interval在 lo，hi里。没加上

        return intervals[: storeIdx + 1]

九章的一个solution

class Solution {
    /**
     * @param intervals, a collection of intervals
     * @return: A new sorted interval list.
     */
    public List<Interval> merge(List<Interval> intervals) {
        // write your code here
        List<Interval> ans = new ArrayList<>();

        intervals.sort(Comparator.comparing(i -> i.start));  //lambda 匿名函数：输入i  返回i.start

        Interval last = null;
        for (Interval item : intervals) {
            if (last == null || last.end < item.start) {
                ans.add(item);
                last = item;
            } else {
                last.end = Math.max(last.end, item.end); // Modify the element already in list
            }
        }
        return ans;
    }
}