56. Merge Intervals Leetcode

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].

这道题目做法有很多中，有用stack的也有直接计算的。这里我采用直接计算的方法。

1.将intervals按照start进行排序

2.然后比较start 如果有 pre.start<=post.start<=pre.end 那pre.end=max(pre.end,post.end) 否者的话就不用进行merge

因为要进行排序，所以时间复杂度为O(nlogn+n)= O(nlogn)

We need to sort the intervals based on start, then we compare pre.start with post.start and pre.end. If post.start fall in [pre.start, pre.end] we need to merge the two based on 

max(pre.end,post.end)

otherwise, we add the interval to the solution directly.

The time complexity in this problem is O(nlogn)

# 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 Interval
    # @return a list of Interval
    def merge(self, intervals):
        solution=[]
        intervals.sort(key=lambda x:x.start)
        for index in range(len(intervals)):
            if solution==[]:
                solution.append(intervals[index])
            else:
                size=len(solution)
                if solution[size-1].start<=intervals[index].start<=solution[size-1].end:
                    solution[size-1].end=max(solution[size-1].end,intervals[index].end)
                else:
                    solution.append(intervals[index])
        return solution