503. Next Greater Element II

Problem:

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.

思路：

Solution (C++):

vector<int> nextGreaterElements(vector<int>& nums) {
    if (nums.empty())  return {};
    int n = nums.size();
    vector<int> res(n, 0);
    
    for (int i = 0; i < n; ++i) {
        int j = i+1;
        for (j; j < i+n; ++j) {
            if (nums[j%n] > nums[i])  { res[i] = nums[j%n]; break; }
        }
        if (j == i+n)  res[i] = -1;
    }
    return res;
}

性能：

Runtime: 404 ms  Memory Usage: 9.8 MB

思路：

Solution (C++):

性能：

Runtime: ms  Memory Usage: MB