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.

這道題是之前那道Next Greater Element I

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;
    }
};

我們可以使用棧來進行優化上面的演算法，我們遍歷兩倍的陣列，然後還是座標i對n取餘，取出數字，如果此時棧不為空，且棧頂元素小於當前數字，說明當前數字就是棧頂元素的右邊第一個較大數，那麼建立二者的對映，並且去除當前棧頂元素，最後如果i小於n，則把i壓入棧。因為res的長度必須是n，超過n的部分我們只是為了給之前棧中的數字找較大值，所以不能壓入棧，參見程式碼如下：

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;
    }
};

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