难度: Medium
原题连接
内容描述
Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.
Return the intersection of these two interval lists.
(Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b. The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].)
Example 1:
Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.
Note:
0 <= A.length < 1000
0 <= B.length < 1000
0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9
思路 1 - 时间复杂度: O(len(A) + len(B))- 空间复杂度: O(1)******
双指针
class Solution:
def intervalIntersection(self, A: 'List[Interval]', B: 'List[Interval]') -> 'List[Interval]':
res = []
i = j = 0
while i < len(A) and j < len(B):
al, ar = A[i].start, A[i].end
bl, br = B[j].start, B[j].end
if ar < br:
i += 1
else:
j += 1
if ar >= bl and br >= al:
res.append(sorted([al, ar, bl, br])[1:-1])
idx = 0
while idx < len(res) - 1:
if res[idx][1] == res[idx + 1][0]:
res[idx:idx + 2] = [[res[idx][0], res[idx + 1][1]]]
idx += 1
return res