leetcode-57. 插入区间

题目

给出一个无重叠的 ，按照区间起始端点排序的区间列表。

在列表中插入一个新的区间，你需要确保列表中的区间仍然有序且不重叠（如果有必要的话，可以合并区间）。

示例 1：

输入：intervals = [[1,3],[6,9]], newInterval = [2,5]
输出：[[1,5],[6,9]]

示例 2：

输入：intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
输出：[[1,2],[3,10],[12,16]]
解释：这是因为新的区间 [4,8] 与 [3,5],[6,7],[8,10] 重叠。

解题思路

Insert + Merge

Like 56. 合并区间, firstly use binary search to find the insert index, then insert, then merge. Time complexity for binary search is $o(\log n)$, for merge is $o(n)$.

Time complexity: $o(n)$
Space complexity: $o(1)$

Merge

Instead of using binary search to find the insert index, we could start by iterate from left to right.

Time complexity: $o(n)$
Space complexity: $o(1)$

Merge2

Keep those intervals that are at the left of the new interval, and those that are at the right of the new interval. Then merge those overlapped.

Time complexity: $o(n)$
Space complexity: $o(n)$

代码

Insert + Merge

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        start_indexs = list(item[0] for item in intervals)
        insert_index = bisect.bisect_left(start_indexs, newInterval[0])
        intervals.insert(insert_index, newInterval)
        # merge intervals
        prev_start, prev_end = intervals[0]
        res = []
        for start, end in intervals[1:]:
            if prev_end >= start:
                prev_end = max(prev_end, end)
            else:
                res.append([prev_start, prev_end])
                prev_start, prev_end = start, end
        res.append([prev_start, prev_end])
        return res

Merge

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        res = []
        i = 0
        while i < len(intervals) and intervals[i][1] < newInterval[0]:
            res.append(intervals[i])
            i += 1
        while i < len(intervals) and newInterval[1] >= intervals[i][0]:
            newInterval = [min(intervals[i][0], newInterval[0]), max(intervals[i][1], newInterval[1])]
            i += 1
        res.append(newInterval)
        return res + intervals[i:]

Merge2

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        left, right = [], []
        new_start, new_end = newInterval
        for i in intervals:
            if i[1] < new_start:
                left.append(i)
            elif i[0] > new_end:
                right.append(i)
            else:
                new_start = min(i[0], new_start)
                new_end = max(i[1], new_end)
        return left + [[new_start, new_end]] + right