[Leetcode]Insert Interval

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].

把一个区间合并到一个连续不重叠的区间集合里，和Merge Interval类似~
把和新的区间有重叠的区间合并为一个大区间即可，其它区间不变。

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

class Solution:
    # @param intervals, a list of Intervals
    # @param newInterval, a Interval
    # @return a list of Interval
    def insert(self, intervals, newInterval):
        if intervals is None or len(intervals) == 0:
            return [newInterval]
        res = []; inserted = False
        for curr in intervals:
            if newInterval.start > curr.end:
                res.append(curr)
            elif curr.start > newInterval.end:
                if not inserted:
                    inserted = True
                    res.append(newInterval)
                res.append(curr)
            else:
                newInterval.start = min(curr.start, newInterval.start)
                newInterval.end = max(curr.end, newInterval.end)
        #remember to add the following procedure
        if len(res) == 0 or newInterval.start > res[-1].end:
            res.append(newInterval)
        return res