## Description

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, output -1 for this number.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.


Note: The length of given array won’t exceed 10000.

## Solutions

### 1. Brute Force

这种有环的操作，记得可以用取余（%）的操作啊！

# Time: O(n^2)
# Space: O(n)
class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
if not nums:
return []
n = len(nums)
res = [-1 for _ in range(n)]
for i in range(n):
j = i + 1
while j % n != i:
if nums[i] < nums[j % n]:
res[i] = nums[j % n]
break
j += 1
return res
# Time Limit Exceeded
# 223/224 cases passed (N/A)


### 2. Stack

可以遍历两次数组，并对下标取模！

# Time: O(n^2)
# Space: O(n)
class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
if not nums:
return []
n = len(nums)
stack = []
res = [-1 for _ in range(n)]
for i in range(2*n):
j = i % n
while stack and nums[stack[-1]] < nums[j]:
res[stack.pop()] = nums[j]
stack.append(j)
return res
# 224/224 cases passed (232 ms)
# Your runtime beats 86.03 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (14.2 MB)