[LeetCode] 986. Interval List Intersections

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mob604757069565 2020-03-05 03:47:00

文章标签 two pointer leetcode line sweep java javascript 文章分类 后端开发

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

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

区间列表的交集。题意是给两组有序的intervals of intervals，请将他们的重叠部分合并成一组。这个题目依然是用到扫描线的思想，同时也用到了追击型的双指针。需要创建两个指针i和j，分别记录interval A和B的位置。试图合并两边的interval的时候，依然采取[start, end]，start两边取大，end两边取小的原则（11,12行）。判断到底是i++还是j++是通过判断当前遍历到的interval谁的end小，谁小谁的指针就++（16 - 20行）。

时间O(n)

空间O(1)

JavaScript实现

1 /**
 2  * @param {number[][]} A
 3  * @param {number[][]} B
 4  * @return {number[][]}
 5  */
 6 var intervalIntersection = function (A, B) {
 7     let i = 0;
 8     let j = 0;
 9     let res = [];
10     while (i < A.length && j < B.length) {
11         let maxStart = Math.max(A[i][0], B[j][0]);
12         let minEnd = Math.min(A[i][1], B[j][1]);
13         if (maxStart <= minEnd) {
14             res.push([maxStart, minEnd]);
15         }
16         if (A[i][1] < B[j][1]) {
17             i++;
18         } else {
19             j++;
20         }
21     }
22     return res;
23 };

Java实现

1 class Solution {
 2     public int[][] intervalIntersection(int[][] A, int[][] B) {
 3         int i = 0;
 4         int j = 0;
 5         List<int[]> res = new ArrayList();
 6         while (i < A.length && j < B.length) {
 7             int low = Math.max(A[i][0], B[j][0]);
 8             int high = Math.min(A[i][1], B[j][1]);
 9             if (low <= high) {
10                 res.add(new int[] { low, high });
11             }
12             if (A[i][1] < B[j][1]) {
13                 i++;
14             } else if (A[i][1] == B[j][1]) {
15                 i++;
16                 j++;
17             } else {
18                 j++;
19             }
20         }
21         // int[][] resArray = new int[res.size()][];
22         return res.toArray(new int[res.size()][]);
23     }
24 }