# 503. Next Greater Element II 下一个更大元素 II

@TOC

## # 题目描述

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.

## # 解题方法

### # 暴力解法

``````class Solution(object):
def nextGreaterElements(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
_len = len(nums)
res = [-1] * _len
for i in xrange(_len):
for j in xrange(i + 1, _len * 2):
if nums[j % _len] > nums[i]:
res[i] = nums[j % _len]
break
return res
``````

### # 单调递减栈

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

C++代码如下：

``````class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
const int N = nums.size();
vector<int> res(N, -1);
stack<int> stack;
for (int i = 0; i < N * 2; i++) {
while (!stack.empty() && nums[stack.top()] < nums[i % N]) {
res[stack.top()] = nums[i % N];
stack.pop();
}
if (i < N)
stack.push(i);
}
return res;
}
};
``````

## # 日期

2018 年 3 月 6 日 2018 年 12 月 28 日 —— 元旦假期到了