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.

// Brute Force
vector<int> nextGreaterElements(vector<int>& nums) { // time: O(n^2); space: O(n)
    int n = nums.size();
    vector<int> res(n, -1);
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (nums[(i + j) % n] > nums[i]) {
                res[i] = nums[(i + j) % n];
                break;
            }
        }
    }
    return res;
}

// Stack
vector<int> nextGreaterElements(vector<int>& nums) { // time: O(n); space: O(n)
    int n = nums.size();
    vector<int> res(n, -1);
    stack<int> st;
    for (int i = 0; i < 2 * n; ++i) {
        int num = nums[i % n];
        while (!st.empty() && nums[st.top()] < num) {
            res[st.top()] = num;
            st.pop();
        }
        if (i < n) st.push(i);
    }
    return res;
}