carloscn
commented
1 year ago Analysis
fn find_common_elements(a:&Vec<i32>, b:&Vec<i32>) -> Vec<i32>
{
let mut ret:Vec<i32> = vec![];
for e in a {
if b.contains(&e) && !ret.contains(&e) {
ret.push(*e);
}
}
return ret;
}
pub fn intersection(nums: Vec<Vec<i32>>) -> Vec<i32>
{
let mut ret:Vec<i32> = vec![];
if nums.is_empty() {
return ret;
}
ret = nums[0].clone();
for i in 1..nums.len() {
ret = find_common_elements(&ret, &nums[i]);
}
return ret;
}
Description
Given a 2D integer array nums where nums[i] is a non-empty array of distinct positive integers, return the list of integers that are present in each array of nums sorted in ascending order.
Example 1:
Input: nums = [[3,1,2,4,5],[1,2,3,4],[3,4,5,6]] Output: [3,4] Explanation: The only integers present in each of nums[0] = [3,1,2,4,5], nums[1] = [1,2,3,4], and nums[2] = [3,4,5,6] are 3 and 4, so we return [3,4].
Example 2:
Input: nums = [[1,2,3],[4,5,6]] Output: [] Explanation: There does not exist any integer present both in nums[0] and nums[1], so we return an empty list [].
Constraints:
1 <= nums.length <= 1000 1 <= sum(nums[i].length) <= 1000 1 <= nums[i][j] <= 1000 All the values of nums[i] are unique.