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
}