Topic description:
You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.
The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
For number 1 in the first array, the next greater number for it in the second array is 3.
For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
For number 2 in the first array, the next greater number for it in the second array is 3.
For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
- All elements in
nums1andnums2are unique. - The length of both
nums1andnums2would not exceed 1000.
Function to be done:
vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums)
illustrate:
1. Given two vectors, such as [3,2] and [2,4,3,1], the first vector is a subset of the second. Request to output the number in the second vector that is larger than an element in the first vector and whose position is also to the right of it. If not found, output -1.
For example, in the example we gave, there is no next greater element of 3, so we output -1. The next greater element of 2 is 4, so 4 is output.
2. After understanding the meaning of the question, the general idea is a double cycle.
We can set the first vector to be vector1 and the second to be vector2.
For each element in vector1, traverse vector2, find its position, and continue looking to the right of its position until a value larger than it is found.
Double loop, this problem can be solved.
code show as below: