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
andnums2
are unique. - All the integers of
nums1
also appear innums2
.
Follow up: Could you find an
O(nums1.length + nums2.length)
solution?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
}