# Next Greater Element II Problem

## Description

LeetCode Problem 503.

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

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, return -1 for this number.

Example 1:

``````1
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Input: nums = [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.
``````

Example 2:

``````1
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Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]
``````

Constraints:

• 1 <= nums.length <= 10^4
• -10^9 <= nums[i] <= 10^9

## Sample C++ Code

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class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
stack<int> st;
int idx, len = nums.size();
vector<int> ans(len, 0);

for (int i = 0; i < 2*len; i ++) {
idx = i % len;
while (!st.empty()) {
if (nums[st.top()] < nums[idx]) {
ans[st.top()] = nums[idx];
st.pop();
} else {
break;
}
}
if (i < len)
st.push(idx);
}
while (!st.empty()) {
ans[st.top()] = -1;
st.pop();
}
return ans;
}
};
``````