Problem Statement

(Source) 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].

Solution

# 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]
        """
        res = []
        n = len(intervals) + 1
        cnt, index = 0, 0
        flag = False
        while cnt < n:
            next = None
            if not flag:
                next = newInterval
                flag = True
                if index < n - 1 and intervals[index].start < newInterval.start:
                    next = intervals[index]
                    index += 1
                    flag = False
            else:
                next = intervals[index]
                index += 1

            if res and next.start <= res[-1].end:
                res[-1].end = max(res[-1].end, next.end)
            else:
                res.append(next)

            cnt += 1
        return res         

Complexity analysis

• Time complexity: $O(n)$.
• Space complexity: $O(n)$.

最后

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