carloscn
commented
1 year ago Analysis
pub fn get_common(nums1: Vec<i32>, nums2: Vec<i32>) -> i32
{
let mut ret:i32 = -1;
if nums1.is_empty() || nums2.is_empty() {
return ret;
}
let mut i:usize = 0;
let mut j:usize = 0;
while i < nums1.len() && j < nums2.len() {
if nums1[i] == nums2[j] {
ret = nums1[i];
break;
} else if nums1[i] < nums2[j] {
i += 1;
} else {
j += 1;
}
}
return ret;
}
Description
Given two integer arrays nums1 and nums2, sorted in non-decreasing order, return the minimum integer common to both arrays. If there is no common integer amongst nums1 and nums2, return -1.
Note that an integer is said to be common to nums1 and nums2 if both arrays have at least one occurrence of that integer.
Example 1:
Input: nums1 = [1,2,3], nums2 = [2,4] Output: 2 Explanation: The smallest element common to both arrays is 2, so we return 2.
Example 2:
Input: nums1 = [1,2,3,6], nums2 = [2,3,4,5] Output: 2 Explanation: There are two common elements in the array 2 and 3 out of which 2 is the smallest, so 2 is returned.
Constraints:
1 <= nums1.length, nums2.length <= 105 1 <= nums1[i], nums2[j] <= 109 Both nums1 and nums2 are sorted in non-decreasing order.