LeetCode #57: Insert Interval

本文介绍了一种在已排序的非重叠区间中插入新区间并进行必要合并的算法。通过两个实例展示了如何处理区间重叠的情况,并给出了具体实现代码及时间复杂度分析。

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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) .
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