503. Next Greater Element II (Medium)

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.

Solution: Stack 99ms

The approach is same as Next Greater Element I
See explanation in my solution to the previous problem
The only difference here is that we use stack to keep the indexes of the decreasing subsequence

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