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.

The Idea: I hashed the locations for the second list, and use that as the starting point of traversing the first list until a greater element is reached.

vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) 
{
    vector<int> next_max_ar;
    next_max_ar.reserve(nums1.size());

    unordered_map<int, int> num_loc;
    num_loc.reserve(nums2.size());

    for (int i = 0; i < nums2.size(); i++)
        num_loc.insert({ nums2[i], i });

    for (int i = 0; i < nums1.size(); i++) {
        int cur_ele = nums1[i];
        int start_loc = num_loc[cur_ele] + 1;
        int next_max = -1;

        for (int j = start_loc; j < nums2.size(); j++) {
            if (nums2[j] > cur_ele) {
                next_max = nums2[j];
                break;
            }
        }
        next_max_ar.push_back(next_max);
    }

    return next_max_ar;
}