interval 系列题

一般遇到interval那就排序吧先，毕竟排好序才好对比时间先后是吧

56. Merge Intervals

Difficulty: Hard

Given a collection of intervals, merge all overlapping intervals.

For example,

Given [1,3],[2,6],[8,10],[15,18],

return [1,6],[8,10],[15,18].

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思路： 按start time排序！如果下一个interval的start 《=上一个interval的end，说明有交集。有交集就合并，直到没交集了就把新的merge好的append到答案集里。

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

class Solution(object):
    def merge(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[Interval]
        """
        res = []
        if not intervals:
            return res
        inter = sorted(intervals, key=lambda x: x.start)
        pre = inter[0]
        for i in range(1, len(inter)):
            cur = inter[i]
            if cur.start <= pre.end:
                pre.end = max(pre.end, cur.end)
            else:
                res.append(pre)
                pre = cur
        res.append(pre)
        return res

57. Insert Interval

Difficulty: Hard

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].

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思路：遍历list找和new的交集，没交集就直接append到res里，有交集就merge更新new，直到new不能更新了就break，res。appen（new）并把之后剩余的list内容加到res里。其实就是找insert和merge的过程。记得用while！不要用for，用for最后容易出问题！

# 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]
        """
        if not intervals:
            return [newInterval]
        res = []
        i = 0
        while i < len(intervals):
            if newInterval.start <= intervals[i].end:
                if newInterval.end < intervals[i].start:
                    break
                newInterval.start = min(intervals[i].start, newInterval.start)
                newInterval.end = max(intervals[i].end, newInterval.end)
            else:
                res.append(intervals[i])
            i += 1
        res.append(newInterval)
        res += intervals[i:]
        return res