986. Interval List Intersections**
https://leetcode.com/problems/interval-list-intersections/
题目描述
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++ 实现 1
类似于将两个 vector 进行 merge. 对于两个区间 p = A[i] 以及 q = B[j], 先求出重叠的部分:
int start = std::max(p[0], q[0]);
int end = std::min(p[1], q[1]);
if (start <= end) res.push_back({start, end});
之后对于 i 和 j 的移动, 需要看哪个区间的末端更近, 移动更近的那个.
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;
}
};
