## 题目描述：

LeetCode 503. Next Greater Element II

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.

## 解题思路：

`时间复杂度O(n)`

https://discuss.leetcode.com/topic/77923/java-10-lines-and-c-12-lines-linear-time-complexity-o-n-with-explanation

## Python代码：

``````class Solution(object):
def nextGreaterElements(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
stack = []
size = len(nums)
ans = [-1] * size
for x in range(size * 2):
i = x % size
while stack and nums[stack[-1]] < nums[i]:
ans[stack.pop()] = nums[i]
stack.append(i)
return ans
``````

`时间复杂度O(n ^ 2)`

## Java代码：

``````public class Solution {
public int[] nextGreaterElements(int[] nums) {
int size = nums.length;
int[] ans = new int[size];
Arrays.fill(ans, -1);
for (int i = 0; i < size; i++) {
for (int j = i + 1; j % size != i; j++) {
if (nums[j % size] > nums[i]) {
ans[i] = nums[j % size];
break;
}
}
}
return ans;
}
}
``````

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