[leetcode] 986. Interval List Intersections

题目

You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, endi] and secondList[j] = [startj, endj]. Each list of intervals is pairwise disjoint and in sorted order.

Return the intersection of these two interval lists.

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 are either empty or represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].

Example 1:

Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]

Example 2:

Input: firstList = [[1,3],[5,9]], secondList = []
Output: []

Constraints:

• 0 <= firstList.length, secondList.length <= 1000
• firstList.length + secondList.length >= 1
• 0 <= starti < endi <= 109
• endi < starti+1
• 0 <= startj < endj <= 109
• endj < startj+1

思路

歪门邪道法：将两个数组合成一个新数组nums并以第一个元素从小到大排序，遍历排序后数组，与此同时记录第二个元素的最大值maxn，当当前第一个元素小于等于maxn时，则表示有重合项，且重合的右边界为min(maxn, nums[i][1])，此时将重合项保存下来，遍历后所有重合项的数组即为结果。

双指针法：双指针分别表示firstList和secondList的索引，建立循环：判断双指针所在区块是否有重合，如果重合则保存结果；随后根据右边界的大小判断移动哪一个指针。直至指针到达数组边界，返回所有保存的结果。

代码

python版本：

class Solution:
    def intervalIntersection(self, firstList: List[List[int]], secondList: List[List[int]]) -> List[List[int]]:
        nums = firstList+secondList
        nums.sort(key=lambda num: num[0])
        res = []
        # print(nums)
        maxn = nums[0][1]
        for i in range(1, len(nums)):
            if nums[i][0] <= maxn:
                res.append([nums[i][0], min(maxn, nums[i][1])])
            maxn = max(maxn, nums[i][1])
        return res

# 双指针方法
class Solution:
    def intervalIntersection(self, firstList: List[List[int]], secondList: List[List[int]]) -> List[List[int]]:
        res = []
        i, j = 0, 0
        while i < len(firstList) and j < len(secondList):
            l1, r1 = firstList[i][0], firstList[i][1]
            l2, r2 = secondList[j][0], secondList[j][1]
            if l1 <= r2 and r1 >= l2:
                res.append([max(l1, l2), min(r1, r2)])
            if r1 >= r2:
                j += 1
            else:
                i += 1
        return res

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