# 496. Next Greater Element I

Easy
The next greater element of some element `x` in an array is the first greater element that is to the right of `x` in the same array.
You are given two distinct 0-indexed integer arrays `nums1` and `nums2`, where `nums1` is a subset of `nums2`.
For each `0 <= i < nums1.length`, find the index `j` such that `nums1[i] == nums2[j]` and determine the next greater element of `nums2[j]` in `nums2`. If there is no next greater element, then the answer for this query is `-1`.
Return an array `ans` of length `nums1.length` such that `ans[i]` is the next greater element as described above.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.
Constraints:
• `1 <= nums1.length <= nums2.length <= 1000`
• `0 <= nums1[i], nums2[i] <= 104`
• All integers in `nums1` and `nums2` are unique.
• All the integers of `nums1` also appear in `nums2`.
Follow up: Could you find an `O(nums1.length + nums2.length)` solution?

### 解題

func nextGreaterElement(nums1 []int, nums2 []int) []int {
stack := make([]int, 0)
m := make(map[int]int)
for _, n := range nums2 {
for (len(stack) != 0 && stack[len(stack) - 1] < n ) {
m[stack[len(stack) - 1]] = n
stack = stack[:len(stack) - 1]
}
stack = append(stack, n)
}
ans := make([]int, 0)
for _, n := range nums1 {
if v, ok := m[n]; ok {
ans = append(ans, v)
} else {
ans = append(ans, -1)
}
}
return ans
}