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.

这道题是之前那道Next Greater Element I的拓展，不同的是，此时数组是一个循环数组，就是说某一个元素的下一个较大值可以在其前面，那么对于循环数组的遍历，为了使下标不超过数组的长度，我们需要对n取余，下面先来看暴力破解的方法，遍历每一个数字，然后对于每一个遍历到的数字，遍历所有其他数字，注意不是遍历到数组末尾，而是通过循环数组遍历其前一个数字，遇到较大值则存入结果res中，并break，再进行下一个数字的遍历，参见代码如下： 

解法一：

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

我们可以使用栈来进行优化上面的算法，我们遍历两倍的数组，然后还是坐标i对n取余，取出数字，如果此时栈不为空，且栈顶元素小于当前数字，说明当前数字就是栈顶元素的右边第一个较大数，那么建立二者的映射，并且去除当前栈顶元素，最后如果i小于n，则把i压入栈。因为res的长度必须是n，超过n的部分我们只是为了给之前栈中的数字找较大值，所以不能压入栈，参见代码如下：

解法二：

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        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;
    }
};

类似题目：

Next Greater Element I 

Next Greater Element III

参考资料：

https://discuss.leetcode.com/topic/77871/short-ac-solution-and-fast-dp-solution-45ms

https://discuss.leetcode.com/topic/77923/java-10-lines-and-c-12-lines-linear-time-complexity-o-n-with-explanation