## 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:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

## Solutions

### 1. Brute Force

最暴力的办法：

# Time: O(nm)
# Space: O(n)
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
n1, n2 = len(nums1), len(nums2)
res = [-1 for _ in range(n1)]
for i in range(n1):
j = 0
while j < n2:
if nums1[i] == nums2[j]:
break
j += 1

for k in range(j+1, n2):
if nums2[k] > nums2[j]:
res[i] = nums2[k]
break
return res
# 17/17 cases passed (320 ms)
# Your runtime beats 5 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


用字典快了一点：

# Time: O(nm)
# Space: O(n)
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
n1, n2 = len(nums1), len(nums2)
dic = collections.defaultdict()
res = [-1 for _ in range(n1)]
for i in range(n2):
dic[nums2[i]] = i
for i in range(n1):
j = dic[nums1[i]]
for k in range(j+1, n2):
if nums2[k] > nums2[j]:
res[i] = nums2[k]
break
return res
# 17/17 cases passed (44 ms)
# Your runtime beats 95.31 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.7 MB)


### 2. Hash Table + Stack

利用哈希表和栈来解决这个问题，哈希表中存下nums2 这个数组中每个数和下一个比之大的数之间的映射。然后遍历 nums1，看对于每一个数在 nums2 中是否有这样的映射，没有的话就是-1.

# Time: O(n)
# Space: O(n)
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
if not nums1:
return []
if not nums2:
return [-1] * len(nums1)
stack = []
greater = {}
res = []
for num in nums2:
while stack and stack[-1] < num:
greater[stack.pop()] = num
stack.append(num)

for num in nums1:
res.append(greater.get(num, -1))
return res
# 17/17 cases passed (64 ms)
# Your runtime beats 38.05 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.6 MB)