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.
题目大意：
给定一个循环数组（末尾元素的下一个元素为起始元素），输出每一个元素的下一个更大的数字（Next Greater Number）。Next Greater Number是指位于某元素右侧，大于该元素，且距离最近的元素。如果不存在这样的元素，则输出-1。
注意：给定数组长度不超过10000。
解题思路：
解法I 栈（Stack）
时间复杂度O(n)
参考LeetCode Discuss：
https://discuss.leetcode.com/topic/77923/java-10-lines-and-c-12-lines-linear-time-complexity-o-n-with-explanation
解法参照LeetCode 496. Next Greater Element I
对于循环数组的处理，将nums数组遍历两次，下标对len(nums)取模
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
解法II 朴素解法
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;
}
}
本文链接：http://bookshadow.com/weblog/2017/02/05/leetcode-next-greater-element-ii/
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