kazi1@的部落格

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
    def insert(self, intervals, newInterval):
        """
        :type intervals: List[Interval]
        :type newInterval: Interval
        :rtype: List[Interval]
        """
        intervals.append(newInterval)
        if len(intervals) < 2:
            return intervals
        intervals.sort(key = lambda x:x.start)
        ret = []
        ret.append(intervals[0])
        for x in xrange(1,len(intervals)):
            y = len(ret) - 1
            if intervals[x].start <= ret[y].end:
                ret[y].end = max(ret[y].end, intervals[x].end)
            elif newInterval.end < intervals[x].start:
                ret.extend(intervals[x:])
                break
            else:
                ret.append(intervals[x])
        return ret