[LeetCode] 986. Interval List Intersections

轉自LeetCode

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: []

Example 3:

Input: firstList = [], secondList = [[4,8],[10,12]]
Output: []

Example 4:

Input: firstList = [[1,7]], secondList = [[3,10]]
Output: [[3,7]]

 

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

Solution

這個是要取兩個list中交集的部分，且這兩個 list 自己裡面的區間沒有重疊，且是排過的

想法如下

因為是取交集，start 會是兩者間比較大的，end 會是兩者之間比較小的

且還要確認 start <= end，才是合法的交集區間

如果有找到，就存到 ans 裡面

那接下是要怎麼移動，檢查當下誰的結束時間比較晚

晚的那個，表示還有機會和其他區間有交集

所以晚的那個不動，移動結束時間比較早的那個

kotlin

	class Solution {
	fun intervalIntersection(firstList: Array<IntArray>, secondList: Array<IntArray>): Array<IntArray> {
	val ans = arrayListOf<IntArray>()
	var i = 0
	var j = 0
	var start = 0
	var end = 0
	while(i < firstList.size && j < secondList.size) {
	start = Math.max(firstList[i][0], secondList[j][0])
	end = Math.min(firstList[i][1], secondList[j][1])
	if(start <= end) {
	ans.add(intArrayOf(start,end))
	}
	if(firstList[i][1] < secondList[j][1]) {
	++i
	} else {
	++j
	}
	}
	return ans.toTypedArray()
	}
	}

view raw interval_list_intersections.kt hosted with ❤ by GitHub