986. Interval List Intersections**

986. Interval List Intersections**

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

題目描述

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.

C++ 實現 1

類似於將兩個 vector 進行 merge. 對於兩個區間 p = A[i] 以及 q = B[j], 先求出重疊的部分:

int start = std::max(p[0], q[0]);
int end = std::min(p[1], q[1]);
if (start <= end) res.push_back({start, end});

之後對於 i 和 j 的移動, 需要看哪個區間的末端更近, 移動更近的那個.

class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        int m = A.size(), n = B.size();
        vector<vector<int>> res;
        int i = 0, j = 0;
        while (i < m && j < n) {
            auto p = A[i], q = B[j];
            int start = std::max(p[0], q[0]);
            int end = std::min(p[1], q[1]);
            if (start <= end) res.push_back({start, end});
            if (p[1] < q[1]) ++ i;
            else ++ j;
        }
        return res;
    }
};