503. Next Greater Element II

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.

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        int sz = nums.size();
        vector<int> res(sz, -1);
        stack<int> st;
        for(int i = 0; i < sz<<1; i++) {
            while(st.size() && nums[st.top()] < nums[i % sz]) {
                res[st.top()] = nums[i % sz];
                st.pop();
            }
            st.push(i % sz);
        }
        return res;
    }
};