# Leetcode: Next Greater Element II

Next Greater Element II Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/next-greater-element-ii
## Basic Idea: Descending stack
##
## Complexity: Time O(n), Space O(n)
class Solution(object):
def nextGreaterElements(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
length = len(nums)
stack = []
res = [None]*length
for i in xrange(length*2):
i = i % length
# if current element is bigger, it's the target of previous undecided elements
while len(stack) != 0 and nums[i] > nums[stack[-1]]:
k = stack.pop()
res[k] = nums[i]
# No need to check if already examined
if res[i] is None:
stack.append(i)

for i in xrange(length):
if res[i] is None: res[i] = -1
return res
```

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