原题

https://leetcode.com/problems/interval-list-intersections/

解法

双指针法, 当A和B的区间不相交时, 有两种情况, A在B后面或者B在A后面, 此时我们需要把处在后面的区间的指针向前进一位. 当A和B相交时, 区间的起点为A, B区间的起点的较大值, 终点为两个区间终点的较小值, 然后将较后面的指针向前进一位即可.

Time: O(m+n)
Space: O(1)

代码

# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution:
    def intervalIntersection(self, A: List[Interval], B: List[Interval]) -> List[Interval]:
        i, j, res = 0, 0, []
        # base case
        if not A or not B: return res
        while i < len(A) and j < len(B):
            # when two interval do not intersect
            if A[i].end < B[j].start:
                i += 1
            elif B[j].end < A[i].start:
                j += 1
            else:
                interval = Interval(max(A[i].start, B[j].start), min(A[i].end, B[j].end))
                res.append(interval)
                if A[i].end < B[j].end:
                    i += 1
                else:
                    j += 1                    
        return res