- Interval List Intersections Medium
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Share 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]]
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：indexA和indexB作为A,B的下标， 1.比较A的right和B的left，A的right<B的left，A和B肯定不相交，indexA++ 2.比较B的right和A的left，B的right<A的left，A和B肯定不相交，indeB++ 3.排除上面两种情况，AB肯定相交，l=max(A[indexA][0],B[indexB][0]) r=min(A[indexA][len(A[indexA])-1],B[indexB][len(B[indexB])-1]),找到相交的首尾，添加到result列表中 代码：python3
class Solution:
def intervalIntersection(self, A: List[List[int]], B: List[List[int]]) -> List[List[int]]:
indexA=0
indexB=0
list=[]
while indexA<len(A) and indexB<len(B):
aleft=A[indexA][0]
aright=A[indexA][len(A[indexA])-1]
bleft=B[indexB][0]
bright=B[indexB][len(B[indexB])-1]
if aright<bleft:
indexA=indexA+1
continue
if bright<aleft:
indexB=indexB+1
continue
l=max(aleft,bleft)
r=min(aright,bright)
if aright>bright:
indexB=indexB+1
else:
indexA=indexA+1
list.append([l,r])
return list
print(list)
if __name__ == '__main__':
print(Solution().intervalIntersection([[0,2],[5,10],[13,23],[24,25]],[[1,5],[8,12],[15,24],[25,26]]))
时间复杂度：O（M+N） 空间复杂度：O（M+N）
