57. Insert Interval【H】【67】

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本文介绍了一个算法问题：如何在一个已排序的不重叠区间列表中插入一个新的区间，并在必要时进行合并。通过两个实例详细解释了该算法的工作原理及实现方式，包括使用lambda表达式进行排序的方法。

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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注释掉的是调用前面一道题的解法的

后面的是看见人家的一个解法的。一个是用lambda演算，一个是直接对最后一个元素进行处理

# 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):

        l = intervals

        l += newInterval,

        li = sorted(l,key = lambda i:i.start)

        #print li

        res = []
        for i in li:
            if res and i.start <= res[-1].end :
                res[-1].end = max(i.end,res[-1].end)
            else:
                res += i,
        return res

        '''

        l = []
        for i in intervals:
            l += [i.start,i.end],

        l += [newInterval.start,newInterval.end],

        l.sort()
        res = []

        i = 0
        while i < (len(l)):
            s = l[i][0]
            e = l[i][1]

            while i < len(l)-1 and l[i+1][0] <= e:
                e = max(l[i+1][1],e)
                i += 1
            res += Interval(s,e),
            i += 1

        return res
        '''
        """
        :type intervals: List[Interval]
        :type newInterval: Interval
        :rtype: List[Interval]
        """