[LeetCode] 986. Interval List Intersections

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

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

/**
 * @param {number[][]} A
 * @param {number[][]} B
 * @return {number[][]}
 */
var intervalIntersection = function (A, B) {
    let i = 0;
    let j = 0;
    let res = [];
    while (i < A.length && j < B.length) {
        let maxStart = Math.max(A[i][0], B[j][0]);
        let minEnd = Math.min(A[i][1], B[j][1]);
        if (maxStart <= minEnd) {
            res.push([maxStart, minEnd]);
        }
        if (A[i][1] < B[j][1]) {
            i++;
        } else {
            j++;
        }
    }
    return res;
};

Java实现

class Solution {
    public int[][] intervalIntersection(int[][] A, int[][] B) {
        int i = 0;
        int j = 0;
        List<int[]> res = new ArrayList();
        while (i < A.length && j < B.length) {
            int low = Math.max(A[i][0], B[j][0]);
            int high = Math.min(A[i][1], B[j][1]);
            if (low <= high) {
                res.add(new int[] { low, high });
            }
            if (A[i][1] < B[j][1]) {
                i++;
            } else if (A[i][1] == B[j][1]) {
                i++;
                j++;
            } else {
                j++;
            }
        }
        // int[][] resArray = new int[res.size()][];
        return res.toArray(new int[res.size()][]);
    }
}

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